®

,. "··

FOUNDATION DESIGN .... HANDBOOK~· ..

:;~-~~~ _ • ~· ,-,.J -,.v-- r - .-1'.1/J'-~1" . -~; • • - ~-.."]C,;:· , . . ~- : ~ • ~ ~ ,.,.,. _ ..... ~ _ ,. ......... • . y ~ -~ / ,.. • ~_. Ill . _ _. - _....,. • . ·- _.....:l . . . IlL" . . _, ... -~ .... . • - - , ·..-· . .:·; _._. ~\.'-.·

Reprinted from HYDROCARBON PROCESSING • Gulf Publishing Company • ©1968 • $1.25 ;i. :V:"'::Ii~·· . "j!l)' .. -- .. <r' . ~, ~~ .... ~ ;~ ~ - -- · -.- - .... · .n:.··~-. oi:ll?·· . . ..::.~·~. ~-. .i;s.: . ..-~·~ .. -~.,.. ; - ~.- .. - --~!1 · 41; : ~:.\.> - -.. ·· -·- ~

This reference manual. has_ b.een reprinted from the reg Jar m issues of HYDROCARBON PROCESSI-NG. Other Handbooks Manuals in the series are: ' .. ··;

... •• o;'

LINES FOR BETTER MANAGEMENT

~· . . . ESS Jf)ISTRUM

FOUNDATION DESIGN HANDBOOK

TABLE OF CONTENTS Page No.

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 Intricate Concrete

4 5

21 27 35

Column Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Unusual Foundation Design For Tall Towers .. . . . . . . .. . . . .. . .. .. . . 44 Foundation Sizing Simplified . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Dowel Sjzing For Tower Foundations . . . . . . . . . . . . . . . . . . . . . . . . . 53 Short Cuts To Tower Foundation Design . .. . .... . . . . . .. .. .. . . .. . . . . 57

VESSEL FOUNDATIONS . . . . . . . . . . . . . . . . . . . . . . . . 60 Foundation Design For 8-Legged Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Pressure Vessel Foundation Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

COMPUTER FOUNDATION DESIGN . . . . . .. . . ... . .. .. . . . . . . .. . . . . . .. .. . 70 How To Calculate Footing Soil Bearing By Computer . . . . . . . . . . . . . . . . . . . 71 Concrete Support Analysis By Computer . . . . . . . . . . . . . . . . . . . . . . 77

FOUNDATIONS ON WEAK SOILS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Foundations On Weak Soils . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Graphs Speed Spread Footing Design . . . . . . . . . . . . . . . . . .. .. . 89 Use Graph To Analyze Pile Supports . . . . . . . . . 93

3

.... , ..

TOWER . ....

FOUNDATION.S '

. . .

,.... .. ..

Foundation Design For Stacks and Towers

The same principles apply in both stacks and towers. Use this method in

making your calculations for either.

V. 0. Marshall, Tennessee Eastman Company, Kingsport, Tenn.

From the viewpoint of the foundation designer, stacks and towers may be divided into two general classifications, depending on the method utilized to maintain them in a ver tical position; (a) Self­supporting, whi ch resist the overturn ing forces by the size, shape and weight of the foundation; (b) Guye d, in which the overturning forces are re­sisted by g uy wires. It is obvious that the con­ditions affecting the design of foundations for these two types will not be the same, an d that it is necessary to treat them separately.

STACKS AND TOW ERS are closely related as far as foundation desi~n is concerned- in fact, the same principks apply. T n the case of stacks, the brick lining is a variable load, corresponding to, and requiring the same trcatment as the li(juid, insulation, etc., in a tower. T his discussion will be based on the design of tower foundations, however, it should be kept in mind that it is also applicable to stacks.

2. Self-Supporting Tower There are two main considerations in designing

the foundation for a self-supporting tower; {a) soil 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 th~ 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.

3. Soil Loading (See Section 20 for complete definition of terms.) The soil loading may be determined by the

following formula:

S=S,+S. (1)

where S =total unit soil loading (lbs./sq. ft.) S. =unit soil loading due to dead load (lbs./sq. ft.) S. = unit soil loading due to overturning m()tllent

(lbs./sq. ft.)

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

w S,= -a- (2)

where a= area of base of foundation (sq. ft.)

W = total weight on soil (pounds) calculated by the following equation:

W= Wc+W. (3) We= Minimum dead load (pounds), which is the

weight of t he empty tower plus the weight of the foundation, including the earth fill on top of the base.

W. =Weight of auxiliary material and equipment supported by the tower and foundation (pounds), which should include the liquid in the tower, insulation, platforms, piping, etc. (Does not include weight of tower)

Thl1 Is a revised ar91clo which wa• previously published In the Augvst, 1943 luue of PITROLIUM REfiNER. All copies of that Issue, all Nprlnh and all copies of the 1940 Process Handbook, In which tho original article wa1 reproduced, failed to meet the demand for this ongln-r­lng data.

When the author considered tho Nvltlon he extonclecl tho tub(ect to Include actual design of foundation types comononly requiNd In the erection of Nflnery veuels. Tho roawlt Is a thorough s .. dy of a M(ect which continues to hold a forefront potltlon In refinery engineering.

Reprint• will be provided In quantity suttident to Include tho demand thcrt has extended Into construction fields outside of reftnlnt. Price $1.00 per copy.

5

Wind Preaeure Pw

:r:

.. c CJ .. -~ • • a::

Foundation Db Top

I G ra~e ~~'V

Foundation ~ Base

F f"A -·-d

FIGURE l Foundotion for self-supporting tower.

6

5. Overturning Load The overturning load S~ is the result of the over­

turning moment. Under ordinary conditions, the only force tending to overturn the tower is the wind pressure.

The magnitude of the wind pressure is obvi­ously a function of the wind velocity, which varies in different lo<:'alities. In many instances Jaws have been enacted which state the wind velocity or wind pressure to be used for design purposes.

The United States Weather Bureau has pro­posed the following formula:

B p = 0.00430V' (4)

where p = wind pressure on a flat surface (pounds/sq. ft.) B = barometric pressure (inches H.) V = velocity of wind (miles per hour)

For a barometric pressure of 30 inches, the formula becomes :

p =0.004V' (5)

I t has been found that the wind pressure on a cylindrical tower is about 60 percent of tha t on a flat surface. For a cylindrical tower, therefore, formula (5) becomes:

P• = 0.0025V' (6) where

P< = wind pressure on the projected area o£ a cylin­drical tower (pounds per square foot) .

In most localities, a wind velocity of 100 miles per hour is considered the maximum. This gives a pressure of 25 pounds per square foot on the projected area of the tower, which is the figure generally used for design purposes. It should be emphasized, however, that this figure is subject to variation in different localities, and that local laws should not be overlooked in this con nection.

(Note: As a matter of interest, the wind pres­sure on an octagon shaped stack is considered to be 70 percen t of that on a flat surf ace.)

Figure 1 represents a tower, mounted on a con­crete foundation. The wind 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 overturnin g moment which can be cal­culated as follows:

Mr= P. L (7)

where Me= overturning moment about the base of the

foundation (foot pounds) P .. =total wind load (pounds) to be calculated as

follows: P .. = p. D.H (8) L =lever arm of wind load (feet) to be calculated

as follows: H

L= hr+ 2 (9)

Do= diameter of tower measured over insulation (feet)

H = heiJ;rht of tower ( feet) h, =height of foundation (feet)

It should be noted that all dimensions are stated in feet, giving the overturning moment (Mt) in foot pounds. This avoids the use of the excessively large numbers encountered with the usual inch pound units. Care should be taken, however, to

use consistent units, that is foot pounds, in all calculations.

The stress, or load, on the soil resulting from the overturning moment (M1 ) varies from point to point, and the maximum load (S,) can be calcu­lated as follows :

Mr s.=z-where

(10)

Z =section modulus of the base of the found"ation. (Note~ Z to be based on dimensions in feet .)

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

I Z= ­ c

where (ll)

I = moment of inertia of the base of the founda­ti?n (based on dimensions in feet).

c = dtstance from neutral axis of foundation base to point o f maximum stress (feet).

Having calculated (51 ) and (52 ) as explained above, the total soil load under maximum dead load conditions can be determined by equation (I).

This maximum soil load occurs at the edge of the foundation, desi~nated as F, and is frequently referred to as the 'toe pressure." It is obvious that the maximum toe pressure (S) should never exceed the safe bearing load of the soil in question.

6. Stability It should be noted that (52 ) is positive at point

F, and negative at E (Figure 1). In other words, the wind load causes compressive stresses on the soil to the left of point A, the maximum compres­sion occurring at F, and tensile stresses of equal magnitude to the right of A, the maximum tension occurring at E.

Since the earth has no strength whatever in tension, it is obvious that the sum of the stresses at any point must be positive. In other words, the base of the foundation must exert a compressive force on the soil over its entire area, otherwise a tensile stress will be produced at E, which means that the tower and foundation will be unstable, and likely to be overturned by the action of the wind.

It was shown by equation (1) that the maximum soil load is equal to (S1 + S2). Since the value of S:~. at point E is negative, the minimum soil load (which obviously occurs at point E) is (S1 - S2).

It is very important to note that the condition of poorest stability occurs immediately after the tower is mounted on the foundation, and before the insulation, platforms, piping, liquid, etc., are in place. In calculating the stability, therefore, (S1 ) must be replaced by (S,m) as follows:

w. S, .. = - .- (2-a)

where S, .. = minimum soil loading due to dead load

(lbs./sq. ft.)

The minimum soil loading which can ever exist, therefore, is found to occur at point E when the dead load is at its minimum value, and can be expressed as fo llows :

Smto = S,,..- St ( 1-a)

Therefvre, in order that (501111) may always be positive, thereby assuring a stable condition at all

times, (S1m) must never be less than (S2). In a perfectly balanced system, (S,m) is exactly equal to (S2 ), in which case

s.. .. = s ... -S.=o (1-d)

Although, such a balanced system is rarely pos­sible, it is the ideal condition. The upward force at E due to the overturning moment is exactly balanced by the dead load, so that the stress at E is zero. The stress at F in such cases is the minimum which can exist and still maintain a stable system.

It should be emphasized that while (S1m) is fre­quently greater than (S,) it should never be less.

It should also be emphasized that the stability is based on the minimum dead load (Wt) while the soil loading is based on the maximum dead load (W).

7. Example No. 1 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 9000 lbs., the maximum wind velocity is 100 miles per hour, and the frost line at the location of the proposed installation is 4 ft. below grade. The maximum safe soil loading is 2000 pounds per square foot.

Solution Since the frost tine is 4 ft. below 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, or 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 mini­mum material better than other shapes. The top course will have a short diameter of 6 ft. since the tower is 4 ft. dia. and allowance must be made for foundation bolts, etc. The short diameter of the base will be assumed to be 13.5 ft. The thickness of the base will depend on the bending and shear­ing forces (see Sections 19 to 19h incl.), however, for the time being the thickness will be assumed to be 2ft.

The weight of the foundation will be calculated as follows (all s lide-rule figures) :

Area of 6 ft. octagon = 0.828 d' = 0.828 X 6' = 29.8 sq. ft. Volume of top course = 4 ft. X 29.8 = 119.2 cu. ft. Area of base (octagon) (a) = 0.828 X 13.5':::.: 151 sq. ft. Volume of base= 2 ft. X 151 = 302 cu. ft. Total volume= 119.2 + 302 = 421.2 cu. ft. Weight of concrete=

421.2 cu. ft. X ISO lbs./cu. ft.= 63,000 lbs. Volume of earth fill

(4 ft. - 1 ft.) X (151 sq. ft.- 29.8 sq. ft.)= 363 cu. ft. Weight of earth tiU = 363 cu. ft. X90 lbs./cu.ft. = 32,700 lbs. Weight of empty tower= 30,000 lbs.

W. = 30,000 + 631000 + 32,700 = 125,700 lbs. W. will be as tollows:

Insulation, platforms, piping, etc. = 9,000 lbs. Water required to fill the tower

( 4 ft . dia.) (50 ft. high) = 39,500 lbs. Total (W.) = 48,500 lbs.

W = 125,700 + 48,500 = 174,200 lbs. (from equation 3) a= lSI sq. ft.

174,200 Jbs. S, = 151 SQ. ft. = 1155 lbs./sq. ft.= Maximum dea~

load on soil (equation 2)

Allowing 3" 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. D.= 4.5 ft. H =54 ft.

P ... = 25 X 4.5 X 54= 6080 lbs. (equation 8) hr= 6ft.

54 • L = 6 + 2=33 ft. (equation 9)

Mr = 6080 X 33 = 200,000 foot pounds (equation 7) Z = 0.1016 d' = 0.1016 X 13.5' = 248.5

From equation (10)

S _ 200,000 ft. pounds b

2- 248.5 = 803 I s./sq. ft.= maxi-mum soil load due to overturning moment.

The total maximum soil load (toe pressure) can be calculated from equation (1) as follows:

S = 1155 + 803 = 1958 lbs./sq. ft.

JAILI 1 llomeftts of o~togonal ....

Short IA!n~h Neutral Alii• Radluaof Diam. Area a of a e to.!ztreme Sec:tloD Gyration r (Feet) (Sq. Ft.) (ll'eet) Fiber c:(Feet) Modulu• Z (Feet)

3 7.46 1.242 1.62 2.74 0.772 3.5 10.6 1.449 1.89 4.37 0.900 4 13.2 1.666 2.16 6.110 1.029

4.6 18.8 1.863 2.43 9.23 1.168 6 20.7 2.070 2.70 12.68 1.286 6.6 26.0 2.277 2.98 16.46 1.415 6 29.8 2.484 3.26 21.90 1.542 6.6 34.8 2.891 3.61 27.90 1.67 7 40.6 2.898 3. 78 34.90 1.8() '1.6 46.6 3.105 4.06 42.80 1.93 8 62.8 3.312 4.33 52.00 2.06 8.6 59.5 3.519 4.59 82.70 2.17 9 66.8 3. 726 4.87 74.10 2.31 9.6 74.6 3.933 6.13 87.30 2.44

10 82.8 4.140 5.41 101.60 2.57 10.6 91.2 4.347 6.67 117.60 2.70 11 100.0 4.554 5.94 136.00 2.83 11.5 109.8 4.761 6.22 164.10 2.00 12 119.5 4.968 6.48 174.50 3.09 12.5 140.0 5.382 7.03 221.00 3.34 13 129.2 6.176 6.76 198.00 3.22 13.6 151.0 5.589 7.30 248.50 3.47 14 162.0 6.796 7.57 277.00 3.60 14.5 174 6.003 7.83 309.20 3.73 15 186 6.210 8.11 342.00 3.86 111.5 199 6.417 8.38 375.00 3.99 16 212 6.624 8·.65 416.00 4.12 16.5 226 6.831 9.02 455.00 4.24 17 240 7.038 9.19 497.00 4.37 17.6 203 7.245 9.46 543.00 4.00 18 268 7.462 9.72 690.00 4.63 18.5 283 7.659 10.00 624.00 4.76 19 299 7.866 10.25 652.00 4.89 19.5 315 8.073 10.55 731.00 5.02 20 332 8.280 10.81 811.00 6.14 20.5 348 8. 41!7 11.08 873.00 5.27 21 365 8.694 11.36 933.00 5.40 21.6 383 8.901 11.62 1005.00 5.63 22 401 9.108 11.90 1085.00 566 22.5 420 9.316 12.17 ll45.00 5.78 23 438 9.522 12.43 1240.00 5.92 2.'!.5 458 9.729 12.71 1320.00 6.04 24 477 9 936 12.98 1400.00 6.17 24.5 497 10.143 13.26 1490.00 6.29 25 618 10.350 13.52 1585.00 6.43 25.6 639 10.667 13.79 1685.00 6.56 26 ti60 10.784 14.07 1787.00 6.68 26.6 582 10.971 14.33 1900.00 6.82 27 603 11. 178 14.61 2010.00 6.94 27.5 526 11.386 14.88 2110.00 7.07 28 6110 11.592 15.15 2220.00 7.20 28.5 6'72 11.799 15.42 2360.00 7.33 29 696 12.006 15.68 2470.00 7.46 29.6 720 12.213 15.96 2600.00 7.58 30 746 12.420 16.23 2740.00 7.71 31 796 12.834 16.77 3021.00 7.97 32 848 13 248 17.31 3330.00 8.23 33 902 13.662 17.86 3660.00 8.48 34 958 14.078 18.38 3980.00 8.75 35 1015 14.~ 18.92 4370.00 9.00 36 1075 14. 19.47 4730.00 9.26 37 1134 16.318 20.01 6130.00 9.51 38 1196 15.732 20.55 5580.00 9.77 39 1260 16.146 21.09 6020.00 10.04 40 1325 16.560 21.65 6500.00 10.28

8

This loading is satisfactory, as the soil will safely support 2000 lbs.jsq. ft.

From equation (2-a)

125,700 lbs. s .... = lSI = 830 lbs./sq. ft.

This is the dead load under the worst stability condition, and since it is greater than the over­turning stress (S~ = 803), the soil below the foun­dation will always be under compression at all points, thus indicating that the foundation is stable.

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

It is interesting to note that the soil loading of 2000 lbs.jsq. ft. allowed in this problem is rather low, as good clay soil will usually support about 4000 lbs.jsq. ft. Care should always be taken, to ascertain the actual load carrying value of the soil at the site of construction.

8. Eccentricity It will 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, acting in a horizontal direction. The combined action of these two forces, that is, their resultant, has thE:. same effect as an eccentric 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 definite relationships between eccentricity and sta­bility, they will be explained as a matter of in­terest.

The eccentricity can be calculated as follows:

e= Me where Wt (12)

e= eccentricity (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 obtained by substituting the proper weight in place of (W1).

It has been shown by previous discussion that the following relationships exist:

Wt s ... =-a-Mt s.=z I

Z=c combining equations (10) and (11)

S :M,c ·=-.­

rearranging equation (12)

Mr= Wte

combining equations (12a) and 13)

S _ Wtec

2- l

(2-a)

(10)

(11)

(13)

(12a)

(14)

It was shown by equation (1-a) that in order to avoid tensile stress at E (which would make the foundation unstable), the maximum value of (S2 ) is as follows:

So= S,., (15)

thus making the value of (Smln) equal to zero, as shown by equation (1-d).

Substitutipg the values obtained by equations (2-a) and (14) in equation (15)

w~ec w. I =-a- (16)

The value of I can be expressed as follows: I= ar• (17)

where r =radius of gyration of base (feet)

substituting in equation (16) w.ec w.

ar' - a (18)

Hence, the maximum value of (e) for stable conditions is

r• e ..... =c

In the case of a circular foundation d

(19)

c=y (20)

Substituting in equation (19) 2r'

e .... = d (21)

Substituting the value of (r2) in equation (21), the maximum value of (e) for a circular base is

(: ). thus confirming the common 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 tower foundations, the maximum allowable eccen­tricity becomes

e ..... = 0.122d (22) The area surrounding the center of the base,

within which the resultant causes a compressive stress over the entire base, is known as the kernel or kern.

It follows, then, that the resultant must always fall within the kern of the base in order to assure stability.

In example No. 1 (Section 7), it was shown that the foundation is stable, since the overturning stress (S2 ) is less than the minimum dead load stress (Smlo).

The stability of this foundation will now be cal­culated (as example No. 2) on the basis of the eccentricity for the purpose of comparing the two methods.

From equation (12) the eccentricity is M t 200,000 foot pounds

e = W, = 125,700 lbs. t.S9 ft.

From equation (22), the maximum permissible eccentricity is

e ... u = O.t22d = 0.122 X 13.5 = 1.64 ft.

Inasmuch as the actual eccentricity (1.59) is less

FIGURE 2

OCTAGOJJ a = o.a2842. c : 0.54ld

I : O.OStid+

Z : O.l016ds

r : o.&5'14

HEXAGOH

a = o.se84' c : o.s7'7ct

I : 0.064•

z : o.104cts

r = o.ae44

a : o.707<l ct•

I :"'"D' · z : o.u8cl, r: o .. aaM

c:..f-1 : o.ot9ct•

z : o.oe8cl'

r=+ Eltmectta of foundotion bases

(Axis A-A)

than the maximum permissible eccentricity (1.64) the foundation is stable, thus confirming the con­clusion reached in Section 7.

9. Method of Calculating SoU Load From Eccentricity

It is possible to calculate the soil loading (toe pressure) as a function of the eccentricity. This method will be explained in order that it may be compared with the method described in Sections 3, 4 and 5.

Let (k) be a factor by which the dead load pres­sure must be multiplied to equal the soil loading due to overturning as follows:

(57)

9

TOW~It

J~----------~----------;k

FIGUU l fiGVRI Ja FIGUU lb

Substituting in equation (1)

S= S;+kS. or

s = s. (1 + k)

From equation (2)

w s.=a-From equation (10)

Mr s.=z-therefore

Mr= S.Z

(58)

(59)

(2)

(10)

(60)

Substituting the value of (S2) from equation (57)

M, = S, kZ (61) therefore

Mr s.=w From equations (2) and (62)

and

W M, S,=-a- = kZ

WkZ M, =-a-

From equation (12)

M,

(62)

(63)

(64)

e= Wt (12)

The value of (e) calculated from equation (12) is maximum for any particular foundation, which is the value governing stability. At the present moment we are concerned with the maximum soil loading (toe pressure) which occurs when the dead load is maximum. It is therefore necessary to substitute (W) in place of (Wt) as follows:

10

Mr e= W

Equation (64) can be written

M, kZ w =--;-

(65)

(66)

Since the term ( ~) occurs m both equations (65) and (66), it follows that

and

kZ e=­a (67)

k-~ - z Substituting this value of

S= St (t+ C:)

(68}

(k) in equation (59)

In the case of an octagonal base,

a= 0.828 d' z = 0.01016 <l'

Substituting these values in equation (68)

8.15e KochJ••=-d-

Therefore, for an octagonal base, equation m~ be written:

( 8.15e) s ....... = s. 1 + -d-

(69)

(70) (71)

(72)

(69)

(73)

For comparison, the maximum soil loading in example No. 1 will be calculated (as example No. 3) on the basis of eccentricity. From previous cal­culations, it was found that:

Mt = 200,000 foot pounds W = 174,200 pounds

Therefore by equation (65)

200,000 s e = 174,200 = 1.1

The maximum soil loading due to the dead load (S1 ) was found to be 1155 pounds per square foot, and (do) is equal to 13.5 feet.

Substituting in equation (13)

( 8.15 X 1.15 ) s = 1155 1 + 13.5

S = 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

....

d

I&

.\;_ / ... . i v. -~~--L!L. --J d

I

0 lq ·-·-· -· lm. m t-· · ---~, _ _......

............

;t:?'.,.. .......

••

. . ' I • . '

-----L-~~---

...-·1 1-·-·-·~~:_j ...... .

_,....- · I _,...... . ~-; ________ _] w

c.

" t

. J ' I q, ____

0

f- ·- ·-· f!.'l·&.. .-::• _, . ~~-,..... ·

' r, "' v

p

k I

...... ......... 1-·-·-·...,.......~ :_j _.-· I .......-· . 1--,: C:::..:.... - . - . - . __j w

t

c

"

FIGUU 3c fiGUU 3d FIGUU le

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 :

S=S.+S• (1)

Since the value of (52) for points to the right of the axis is negative, the value of (S) will be:

S = S, - s. (see equation 1-a) ( 1-b)

Equations (1) and (1-b) can be combined as follows:

S=S,±S, (1-c)

From equation (2) w s.=-a (2)

The value of (52 ) can be written: \Vee'

S, = ar' (see equations 14 and 17) (23)

Thereiore,

S- W +Wee' - a - ar'

Simplifying:

S= ~ (I±~~') (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), (14) and (17) referred to above were used to deter­mine the s tability 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 (Wt) depending on the dead load for which the stress is to be determined. It is

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 in Tower Shell

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

M 1 = P .. ( ~) (26)

where M, = bending moment about base of tower (foot

pounds). also

S _ M,c '- I (27)

where S, = unit stress in tower shell due to bending

moment (Mt). (lbs./sq. ft.)

Note: The unit stress in the tower shell (St) is calculated in pounds per square foot in order to be consistent with the other calculations which are in foot-pound unts. Steel s tresses, 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 necessary to divide (S,) by 144.

T he shell is a hollow cylinder, in which case:

D c=-y (28)

and '71'

I = 64 (D•- D\) (29)

where D = outside diameter of tower (feet) D, = inside diameter of tower (feet)

when

and t = thickness of shell (feet)

D -D,=2t D,=D -2t

(30) (31)

11

Substituting in equation (29)

'1r I= 64 [D'- (D -2t)'] (31)

Substituting the values of I and c in equation (27)

D M, 2 32M,D s. = ___ __; __ _

<rr [0'- (D -2t)'] '11' &f[D'- ( D-2~)' ]

32M,D - <rr [D'- D' + 8ntt -24D't' + 32Dt'-l6t')

32M,D - <rr [8D't-24D't'+ 32Dt' - 16t') (32)

The values of t 2, t• and t4 are quite small and

the three terms in the denominator containing them may be neglected without introducing ap­preciable error. For practical purposes, therefore, equation (32) may be written as follows:

32M,D 5•= 8?TD•t

This equation may be reduced to:

4Mt Sa= 71'D't

(32-a)

(32-b)

T he thickness of shell plate required to resist the bending moment only, is therefore

4Mt t= <rrD'S, (33)

By multiplying the stress in pounds per square foot (S1 ) by the shell thickn~ss (t) the stress per foot of circumference is obtained as follows:

4M, tS. = <rrD' (34)

12. Foundation Bolta for Self-Supporting Tower The foundation or anchor bolts for a self-sup­

porting tower are required to resist the overturn­ing moment (Mr) resulting from the wind pres­sure, after allowance has been made for the resist­ance offered by the weight of the tower. Obviously the I'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 (34). 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 (Db) in place of (D). The stress per foot of bolt circle circumference can then be written :

where

4M• '1rD!,.

Do= diameter of bol t ci rcle (feet).

(35)

The compressive stress per foot of circumfer­ence due to the weight of the tower is

12

w. 11'Db (36)

The maximum tensile stress per foot of circum­ference to be resisted by anchor bolts is

4M, W. <rrD•' - ?TDo (37)

As5uming that the number of bolts is repre­sented by (N), each bolt will be required to carry the stress over a portion of the circumference rep­resented as follows:

71'D. ---w- (38)

The load to be carried by each bolt can be expressed

S.. _ 'lTD• .( 4Mt W. ) - N <rrD•'- <rrD!,

4M, W . = NDb - --w- (39)

where S.. =maximum load on each bolt (pounds).

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

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 root of the thread. An additional allowance, usually %", should be made for corrosion.

The number of foundation bolts should never be tess than 8, and should preferably be 12 or more, as the larger the number of bolts, 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 wi!l 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 an­chor.ing the bolt in the concrete (see Figure 7).

13. Guyed Towers In cases where the tower is very high, it is

sometimes found desirable to maintain stability by means of guy wires rather than a large founda­tion. 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 used for 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 at a point ap· proximately 2/3 (sometimes ~) of the tower height above the foundation.

14. Pull on Guy Wires The maximum 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.

The pull on the guy wire can be expressed as follows:

R. R,= Sine (40)

or R, = R. esc 8 (41)

where R,= pull on guy wire due to wind pressure

(pounds) R. =horizontal wind reaction at collar (pounds)

8 = angle that the guy makes with the vertical (degrees)

The value of the angle 8 will usually lie between 30 and 75 degrees.

The vertical component of the pull on the guy wire can be expressed in any of the following ways:

R, X cos fl R. X cos 8

Sin 8 R. X cot 8

(42)

(43) (44)

It is important to note, however, that there is always some initial tension on the guys which must be considered. This initial tension may be assumed to be 5000 lbs.jsq. in., which amounts to 1000 lbs. for )12" wires and 250 lbs. for }4" wires. The weight of the wires may be neglected, when using tht:se figures.

The actual vertical component will be a func­tion of the total pull on the guy wire, which is

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

Rv= (R, + Rt) cos 9 (45) where

Rv =vertical component of pull on guy wire (pounds)

Rt =initial tension on wire (pounds).

The value of the reaction at the collar (Rc) may be determined by calculating the moments about the base of the tower (the top of the foundation).

The wind moment was found by equation (26) to be

p,. (~) The resrstmg mom~nt arm at the collar is h11

therefore the reaction (Rc) may be calculated as follows:

or

R.= p,. ( 1f)

h,

P,.H R.= 2h

I

where

(47)

(48)

h, =height from top of foundation to collar (feet}.

15. Foundations for Guyed Towers It was shown by equation (1) that the total soil

loading, to be considered in the design of tower foundations, is the sum of (S1 ) which is the dead load, and {52 ) which is the load due to the over· turning, or wind moment. In the case of the guyed towers, there is no overturning moment, however, the wind pressure does have an important effect on the foundation, as the soil is required to resist the vertical component of the pull on the guy wires.

For guyed towers, therefore, equation (1) must be revised as follows:

S=S.+s. (49) where

5 1 = unit soil loading due to the pull on the guy wire. (pounds/sq. ft.)

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

S-R. r- a

From equation (2)

S.= w a

Substituting in equation (49)

S= W+R. a

16. Foundation Bolts for Guyed Towers

(SO)

(2)

(51)

The foundation bolts for guyed towers are re­quired to resist the shearing action 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 the horizontal reaction to the wind pressure at that point. This is equal to the difference between the total wind

13

pressure and the reaction at the collar and can be expressed as follows:

L=~-L (~) where

R- = horizontal wind reaction, or shear, at the base of the tower. (pounds)

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

duces a negative bending moment at the collar,

'1 ... _ _ ._ ........ _.a."-'- :or -t ~--..

' T .__ ;!_-' --""*" .:-. b -It'--lr- ..... . r-~-1

1 t41 ,r--,. ~--- T--- -H--r ----- -------I 1 I

1 tl.::~ I ~--------i--=- ·t---------------

I I I

FIGURE 5

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

M.=- ;i£ (H - h,)' (53)

P,.H ( H )' .M, = -2- 1 - 2h, (54)

where Me= negative bending moment at collar. (foot

pounds) MP = maximum positive bending moment between

collar and ba~e. (foot pounds).

Having determined the bending moments, the stress in a given shell, or the shell thickness re­quired to resist the bending moment may be cal­culated by substituting the value of (M.) or (Mp) in place of (Mt) in equations (32-b) and (33).

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 s6il. 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 12" for longer piles.

Wooden piles generally depend on the frictiona l 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 govern 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 common method of calculating the safe load is by means of what is known as the "Engineering News Formula," which

14

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

For drop hamme1·

p- 2W .. f - P• + 1.0 (55)

For steam hammer 2W.,.f

P= P•+O.I (56) where

P = safe load which each pile will support. (pounds)

w .. = weight of hammer. (pounds) f =height of hammer fall. (feet)

p .. = penetration or sinking under the last blow, on sound wood. (inches)

Care should be exercised in driving piles, to assure that they are deep enough to develop their full strength, but they should not be driven too much, as this practice results in splitting or break­ing, 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 than 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 belqw the water level, otherwise they will decay rapidly.

The reinforced concrete cap is constructed on top of the piles in such a mann~r that the piles extend about 6" into the concrete (see Figure 6) .

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 loading, it becomes necessary to calculate the stresses in th~ foundation itself, to see that they do not exceed the allowable limits.

The first step in this procedure is to determine

FIGUR£ 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 (S1) is represented by the rectangle (jklm). The wind load (52 ) , which is positive on one side of the centerline, is indicated by the triangle (mpw). On the opposite side of

the centerline the wind load is negative, thereby counteracting a portion of the dead load (wlc). The actual soil loading will therefore be repre­sented by the area (jkcp). However, the. weight of the base, and of the earth fill above the base (area jkno Figure 3b) do not exert any upward force on the foundation, and may therefore be deducted from the total load, for the present purpose. The effective upward reaction will then be the area ( oc1 p) in Figure. 3b.

19a. Diagonal Tension The vertical shear, resulting from the upward

reaction of the soil, produces diagonal tension 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 indicated by pt>int (Z,) in Figure 3c. In other words, the foundation tends to break along line (ZZ1). The vertical shear to be resisted is equal to the net soil pressure on the part of the foundation outside the critical section.

For design purposes, therefore, the load will be the area (oqrp), (Figure 3c) applied over the area (a, b1 fg), (Figure 3). Because of the irregu­lar shape of the load diagram, its magnitude can be more conveniently calculated by breaking it up into its component parts, the total load (V ,) being the sum of the individual loads, as follows:

Shape of ~rt Rectangular Prism Wedge Wedge Wedge Pyramid Pyramid

Outline hi plan (FII'. 3)

a, b, u, t, a, t, g b, fu,

a, b, u, t, a, t, g b, fu, ·

Outline In el nat loo (Fl ... 3e)

oqrv oqrv O<lrv

rvp rvp rvp

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

v. fo= b'jd, (80)

where . f• =unit stress in concrete (in diagonal tenston)

due to vertical shear load. (pounds/sq. in.) V • = vertical shear load, outside the critical section

(see Figures 3 and Jc). (pounds) b' = width of critical section which serves to resist

diagonal tension stresses (line a, b, Figure 3). (inches)

j = ratio of lever arm o£ resisting couple to depth (dr) (see Table 2).

dr = effective depth of base measured from top of base to centerline of reinforcing steel. (inches)

Example 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 till

Total

Unit soil loading

63,000 lbs. 32,700 lbs. 95,700 lbs. 95,700 lbs. 633 lb I f = 151 f = S. SQ. t . SQ. t.

Total unit dead load (S,) (jm, figure Jc) = 1,1551bsJsq. ft. Unit dead load due to weight of base

and earth fill (jo) = ~ Net soil load (om) = 5221bs./sQ. ft.

Maximum unit wind load (S.}, (mp) = 803 Maximum effective unit shear load (op) 1,3251bs./sq. ft.

162 Line {mw) =z-=81"

d,=72". d,=2lw. d.=I62" d.=72±21 ±21 = 114" Line (gf) =67.1" (see Table 1).

114 Line (m, w) =z-= 57".

. 67.1 X 57 _ 47 2, _ (b') Lsne (a, b,) = 81 - . - .

. 67.1 - 47.2 - 995" Ltne (gt,) = 2 - .

, 162- 114 11

Ltne (a, t,) = 2 = 24

Factor j = .87 (see Table 2). 803 X 57

Load (m, r) = 81 = 5651bs./sq. ft. Load (qr) = 565 ± 522 = 1,087lbs./sq. ft.

Calculate shear load (V, )

7 2" 2 w X 1,087 lbs./sq. ft.

4 · X 4 144 SQ. in. - 8,550 lbs.

9.95" X 24" X ~~~7 238

47.2 X 24 X 2 X 144 9.95 X 238 X 24 X 2

3 X 144

1,805

= 935

263 Total (V,) = 11,558 lbs.

Calculate unit stress m concrete (equation 80)

11,558 lb I . fo = 47.2 X .87 X 21 = 13.4 s. sq. tn.

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

19b. Depth 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-Z2), (Figure 3c) at the edge of the pedestal. This shearing load may be determined as follows:

S.= s.t s. (81)

The stresses in this case are not distributed over the foundation area, but are concentrated at the edge of the pedestal.

Then

S, = total maximum unit shearing load, (lbs. per lineal foot of pedestal perimeter).

S, = unit shearing load due to dead load. (lbs. per lineal foot of pedestal perimeter).

S, = maximum unit shearing load due to overturn­ing moment. {lbs. per lineal foot of pedestal perimeter).

The value of (S.) can be determined by adding the weight supported by the pedestal to the weight of the pedestal itself, subtracting the load· carrying value of the soil directly under the pedestal, and dividing the difference by the pe· rimeter of the pedestal base as follows :

W, ± W, ± WP- (a, S. u) S, = L, (82)

where W, =weight of foundation pedestal (top course).

(pounds) a, = plan area of foundation pedestal. (sq. ft.)

S at> = maximum allowable unit soil loading. (pounds/sq. ft .)

L, = perimeter of foundation pedestal. (feet)

Obviously, if the value of (ap S .. u) is equal to

15

or greater than (W. + W .. + Wp), the value of (S,) becomes zero, and (S1) will then be equal to (S-6).

The value of (S6 ) 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). T~e bolt circle was assumed to be a hollow cylinder, 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 (Mr) in place of (M1), and appropriate values of (I) and (c) in equation (27), depending on the shape of the pedestal.

Reduced to their simplest forms, the equations for the ordinary foundation shapes are as follows:

Octagon Mr

5•'= .814dp1

Hexagon Mr

s. = .832d.' Square

Mr s.= .943d:

Circle Mr

s.=TBW

(83)

(83a)

(83b)

(83c)

In these equations (d11) is the short diameter of the pedestal (feet) .

Once the shearing load (S,) per foot of pedestal perimeter is known, i~ is a simple matter .t'! ~al­culate the unit stress m the concrete, by dtvtdtng (S,) by tpe effective depth of the base, as follows:

s. f,= 12dr (84)

where . f, = unit stress in concrete base due to punchtng

shear. (pounds/sq. in.)

Note: The factor 12 is introduced for the pur­pose of convertin~ (51 ) from (pounds/lin. foot) to (pounds/lin. inch) as unit stress (fr>) is in terms of (pounds/sq. in.).

Example No. 5. To illustrate the procedure, the punching shear will be calculated for the founda­tion considered in example No. 1.

Calculate dead load shear (S•) by equation (82)

W. = 30,000 lbs. W.= 119.2cu. ft. X 150lbs. a.= 29.8 sq. ft.

w. = 48,500 lbs. = I7,850lbs.

s.w= 2,000 lbs./sq. ft. L. = 2.484 X 8 = 19.87 ft.

3(),000 ± 48,500 + 17,850- (29.8 X 2,000) S.= 19.87

= 1.850 lbs.llin. foot.

Calculate shearing stress due to overturning moment (S1 ) by equation (83)

16

M. = 200,000 foot pounds (see section 7). d."= fl=36 ..

Sa= ::~~6 = 6,820 pounds/lin. foot S, = 1,850 ± 6,820 = 8,670 pounds/lin. foot. d.= 21"

8,670 44 d I . f, = 12 X 21 :....: 3 . poun s sq. an.

(81)

(84)

This stress is satisfactory, as 120 pounds/sq. in. is permissible. (See T able 2.)

In the case of guyed towers, or stacks, the shear load due to overturning moment (S~) does not apply, but is replaced by

( R;/L,) which is the load due to the pull on the guy wires, as f~lows:

R:r Sl(n,t() = S. + -y:;-

(81a)

19c. 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 fill 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 q1 r 1 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 dia­gram, it is difficul t 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 (Mb) 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:

OatllDe m,t-a• OatlJDe In eleva-Fl ... tloa, FQ-. 3e Lever Ann

Rectangular abut

Distance (at) Prism q,r,v,o 2

Distance (at) X 2 Wedge atg q, r,v,o 3

Wedge bfu q,r,v, o Distance (at) X 2

3

Wedge abut r,v, p Distance (at) X 2

3

Pyramid atg r sVs P Distance (at) X 2

3

bfu Distance (at) X 2

Pyramid r,v, p 3

In calculating the amount of re~nforcement re­!}Uired, it is assumed that the portiOn of the. base designated (abut), (Figure 3d) acts as~ canttl7ver beam (of rectangular cross-section) havmg a wtdth equal to one face of the pedestal (a b), a depth equal to the effective depth of the base ( d r) and a length equal to (at).

Having calculated the bf'nding moment as pro­posed above, the next step is to check the depth

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 moments in terms of inch-pounds, as the stresses in concrete and steel are usually given as pounds per square inch, whereas in figuring soil loading foot-pound units are used, as soil loading is usually stated as pounds per square foot.)

For balanced design, that is, conditions in which both concrete and steel are stressed to their full allowable capacity, the required depth (de) of the base may be determined as follows:

_/ Mb dr = "' f 'b • P• J •

where · (85)

dr = depth of base, measured from top of concrete to centerline of reinforcing steel: (inches)

Mb =bending moment in base. (inch-pounds) f. = safe working stress, reinforcing steel in ten­

sion. (pounds per sq. in.)

P• = ( A0• dr ) = ratio of effective area of rein­forcing steel to effective area of concrete.

j = ratio of lever arm of resisting couple to depth (dr).

b.= width of beam (line ab, Figure 3d). (inches) A.= effective cross sectional area of steel reinforce­

ment in tension. (square inches)

If the design is balanced, that is, the actual depth of the base ( dr) is that calculated by equa­tion (85), the value of (A,) may be determined 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:

Mo A.=~d Io' t

(87)

If the depth ( dr) is less than required by equa­tion (85), it is recommended that the dimensions 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 necessary to increase the amount of reinforcement used. The determination of the amount of rein­forcement required for such special cases is be­yond the scope of this article, and reference is made to the various publications dealing speci­fically with concrete design for further details.

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 is recommended that the center-to-center distance be about 4 inches if possible, but not less than 2~ times the bar diameter for round bars, or 3 times the side di­mension for square bars. Generally speaking, a large number of small bars (0. %. or ~ inch) are preferable to a smaller number of larger bars.

It should be borne in mind that 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

-----------------·-··- . -

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), using 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 of reinforcing bars parallel to each of the axes, i.e., four sets of bars for an octagonal base, three sets for 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.

Example No. 6. Determine bottom reinforcement for the foundation referred to in example No.1.

Figure bending loads 72"

Line (m2 w) =-2- = 36"

803 X 36 Load (m. r,) 81 = 357 pounds/sq. ft. Load (q, r,) = 357 + 522 = 879 pounds/sq. ft. Load ( v, p) = 803 - 357 = 446 pounds/ sq. ft.

Figure moment (Mb)

Line (ab) =29.8" Line (ta) = 45" • Line (gt) = 18.65" 29.8" X 45" 45" . 144 · X 879 pounds/sq. ft. X - 2- = 184,000 m.-lbs sq. m.

18.65 X 45 879

45 X 2 144 X X 3 = 153,000

29.8 X 45 446 X 45 X 2 144 X -l- --3- = 62,300

18.65 X 446 X 45 2

45 X 2 144X3 X x-3- = 51,900

Total (Mb) = 451,200 in.-lbs.

Check depth of base for balanced design ( equa­tion (85)

f.= 18,000 p.= .0089 j = .87

b.= 29.8" f. p. j = 138.7

d _/ 451,200 -- 10.5" f(baloa .. •l ="' 138.7 X 29.8

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).

451,200 . A.= 18,000 X .87 X 21 = 1.37 sq. 10•

Use 0 inch deformed square bars ( .25 sq. in. area).

Number required ~i3J = 5.5. Use 6 bars within the width of beam (ab).

Spacing 29·t' = 4.96''. Use 5-inch spacing en­tirely across side (gf). which will require 67·t' = 13 bars per set. Four sets of bottom reinforcing bars will be required for the octagonal foundation.

17

19d. Reinforcement to Resist Stresses Due to Uplift

As explained previously, the wind moment cre­ates a positive soil load on one side of the center­line, and a negative load on the opposite side. In other words, the action of the wind tends to lift the foundation on the negative side. This upward force, or "uplift" effect, is resisted by the weight of the concrete base itself, and by the weight of the earth fill on top of the base. It therefore be­comes necessary to reinforce the top of the base, to resist the resulting negative bending moment.

The procedure is quite similar to that described for the upward soil reaction (Seetion 19c). The load acts on the area (abfg), and the outline of the theoretical beam carrying the load ig (abut) as in Section 19c. However, in this case the load is th~ weight per square foot of concrete base, plus the weight per square foot of the earth fill, and is uni­formly distributed, thus simplifying the calcula­tions. After figuring the moments, the reinforce­ment is determined in exactly the same manner as explained in Section 19c, using the equation

M .. A.=-f 'd o1 t (87a)

In this case, ( dr) is the depth of the base fron: the centerline of the upper layer of reinforcement to the bottom of the base, and (Mu) is the bending moment due to the uplift forces (inch-pounds).

Example No. 7. Determine top reinforcement to resist uplift in the foundation referred to in ex­ample No.1.

Weight of concrete 1 SO lbs./ cu. ft. X 2 ft.

Weight of earth 90 lbs./ cu. ft. X 3 ft.

Total

= 300lbs./sq. ft.

= 270 lbs./sq. ft. = 5701bs./sq. ft.

TAILI 2 Conttcaftta Applying to foundation DNI8"

Mt.ture: Cement .............................. I 1 Sand ................................. l l Coane Aallftaate ..................... 4 5 ---

fb Safe bcvlna load on concrete (lbs./IIQ. in.). 500 376 ---f~ Ultimate compreuive etrength (lbt./IIQ. in.) 2,000 1.500 ---fo Safe unit atress In extn:me fibl!t' of concrete

(in comprcsaion)~(Jba./sq. ln.) ......•.. 800 600 ---f4 Safe unit strass In concrete due to vertical

shear (diaronal tenaion} (lbs./~q. in,) •• 40 30 ---f. Safe unit stresa in concrete baae due to

punchina shear. (lb&./IIQ. in.) ••........ 120 90 ---f. Safe workinll, atreu. steel reinforcement in

tet~tion. (I e./tq. in.) .................. 18.000 18.000 ---(f. j) (.l.bch-pounds) .......................... 16.600 16.000

(1. Jp.) (l11ch-pounde) .......................... 138.7 88.9 ---I Ratio. lever arm of rnlatlng couple to

depth (dr) •••.•••.••••..•••.••.••.•.. .87 .89 ---[ n .. Efu.. I Ratio, modulue o( elatticity of steel to that

of concrete .......................... 16 IS ---{p.•Ao/ I

/b.dr Ratio. effeetlve area of tenalon reinforce-

rnent to elf ~tive area of concrete ...... .0089 .0056 ---•u Safe bond ttrea.t (concrete to steel rein·

forcemeDt} per "Ualt of area of ~urface

of ~:in(f:~~~·! ...•••................ 60 45 Deformed bart ..................... 76 56

• Theae tiguree may be allihtly lno:r~ by makin& "U"-beoda on thf' ende of tbe relnfordnl ban.

Note:-The 1:2:4 mixture Ia te«>mmfJlded u moet satiafactory for founda­tloru of the type. The conetanta for tha 1:2:5 mixture are preaented aa a matter of latereet.

18

Moment 29.8 X 45

570 .~

144 ·X X 2 = 119,000 in.-lbs.

18.6:: 45 X 570 X 45 ~ Z = 99,500

Total (M~) ;:::: 218,500in.-lbs.

From equation (87a)

A _ 218,500 inch lbs. • - 18,000 X .87 X 22 .636 sq. in. within beam

width (29.8")

Use 0-inch deformed square bars, at to-inch centers.

19e. Bond In order for the reinforcement to be effective,

the strength of the bond between concrete and steel must be sufficient to permit the reinforce­ment to develop its full strength. The bond stress may be calculated by means of the following formula:

v. u = :t. jdt- (88)

where u = bond stress per unit of area of surface of bar.

(pounds) ::t. = sum of perimeters of bars within the limits or

the beam width (ab). (inches)

Example No. 8. Check bond stresses in example No.1.

Bottom reinforcement

v. = 11,578 lbs. (See example No. 4) :t.= 6X .5 X 4= 12"'

By equation (88)

11,578 u = -t'""z'"'x...,..;.:.8:::7_,.X....,...,..2t=- =53 lbs.

Bond stress for bottom reinforcement is satis­factory, as 75 pounds is permissible (see Table 2).

Top reinforcement

Figure shear 47.2" X 24"

144 X 570 lbs.

9.95 X 24 X 570 144

Total (V.) ::t. = 3 X .5 X 4 = 6

5427 u = 6 X .87 X 22 =48tbs.

=4,480Jbs.

= 947

= 5,4271bs.

(88)

The bond stress in top reinforcement is satis­factory, as 75 pounds would be atlowed.

19f. Bearing Stresses The bearing stresses (where the steel tower

rests on the concrete pedestal) seldom cause any difficulty, but should be checked as a safety pre­caution. The bearing stresses consist of the stress due to wind pressure, plus the stress due to the dead load as follows :

Bearing stress= 4M,/rrD,2 + (W. + W.)/17'0, (37a)

(See Sections 11 and 12.) Equation (37a) gives the bearing stress in

pounds per lineal foot of shell circumference. These stresses are spread over the area of the base ring, therfore for practical purposes the unit bear­ing stress can be determined as follows:

4M .. + w.+w. 'IT D.' 71' D,

f. = 12r ... in which

r., = width of the tower base ring. (inches) fb = unit compre~sion stress on concrete.

(pounds/sq. 1n.)

(37b)

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

FIGURE 7

of cases it mav be used in the above form with reasonable accuracy.

For guyed towers, equation (37b) becomes:

4M, + R.+w.+w. 'iT n,• 71' n,

fb = -- 12r., (37c)

19g. Allowable Stresses in Foundation lt 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 5tresses for diagonal tension, punching shear, bending (upward and downward) and bond in the reinforcement are well below the allowable values. As the examples in

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. Sugge.tions and Recommendations The calculations explained above provide for

reinforcement to resist the stresses due to the 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 !'lab 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 as a guide.

In the case of very large foundations, consider­able 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 screwed into sleeve nuts from the top.

Nomenclature

Ao - effective cross sectional ~rca of steel reinlorce~t iu Len· sion (sq~arc inchu)

For balanced dul rn Ao - bo dr Pr (86)

If depth (dr) is anater than required by cq,ua1ion (85)

Mb A o - T.J(jj • (87)

I' or top reinfo rcement of slab to resist uplift struses:

M. 8' ) Ao "" f,jdr (ta

a~ area of base of foundation (sq. ft.) &v = plan area of foundation pcdcatal (sq. ft.) B = barometric prcuure (inches Ha)

bo = width of the critical section (equal to the width of tbe lace of the pedestal) anumed to act as a cantilever beam resist· inr the bending streuu (line ab, Fi1urc 3d) (inches)

b' ~ width of critical aection which serves to resist the diagonal tension stresses. ( line &1 bs, Figure 3.) (inches)

c - distance from neutral axi s of foundation b"<e to point of maximum st rus. (feet)

c' = di5tance from centro idal a•is of foundation base to any point under conoidcration. ( feet}

D ,.. outside diameter of tower. (lett)

D, = inside diameter of tower. (feet)

Do= tower diameter measured over insulation. (feet)

Db- diameter of foundation bolt circle. (feel) do ~ short dial1)oter of foundation bue. (feet)

d. = shol'l diameter of critical section Cor diagonal tension stresses (sec: Fiaurea 3 and Jc). (inclu:&)

dt = eUective depth of base of foundation, measured from top of base 10 centerline of reinforcin&' steel. (inche&)

19

d, = short diameter of found ation pedestal. (feet) Ec ... modulus of elasticity of concrete .. Eo - modulua of elasticity of reinforcinr steeL

e = ecctntricity. (feet) This factor i1 the distance from the centroidal axit of the foundation to the r,oinf at which the resultant of the dead load and the wind oad intersects the base of the foundation. The eccentricity can be calcullled as followJ :

Mr e = l jj;- (12)

Equation (12) .rives the eccentricity at tbt condition of poorest atability, that is1 with the minimum dud load. This 11 the value which ordmarily is used for design purposes, however, it ia obvious that the eccentricity for maximum dead lo11d c~nditions can be .calculated by substituting the value of (W) in equation (12) in place of (Wo). The m4.n"m•m value which it is pos&iblt for (e) to han and still maintain tbe stability of the foundation is

•' e .... --c-' or (19)

z ema.:- - -a

(19a)

Values of (emu) for various foundation shapes are as follows:

Ocl&Jon,: e .... - O.l22d Hexagon: e .... - 0.121d Square: e •• ,- O. ll8d

Circle: d emaa--

8-

(22) (22a) (22b)

(22c)

The value of (e) u calculated by equation ( 12), and bas~d on the minimum dead load (W •) should 11cvcr oxcoed the value calculated by equations (19) or (19a).

F - barometric prusu.re. (inchu H 1)

I - height of hammer fall. (feet) 1, = unit bearing stress on concrete. (pounds/sq. in.). (See equa·

tions 37b and 37c) fe - 1;8.ff! unh streu in ~xtreme fiber of concrete (In compres·

sion). (pounds/sq. in.) f•'- ultimate compt·euive ttrength of concrete. (pounds/sq. ~n.) r.- unit stress in concrete (in diaronal tension) due to vertocal

shur load. (pounds/sq. in.) 1--unit streaa in concrete base due to punching shear. (pounds/ sq. in.)

s. fv - i2dt (M)

fo- safe workin1 strtss, ateel reinforcement in tension. (rounds/ sq. in.)

H - height of tower. (feet) h r - keigbl of foundation. (feet) hs = heilht of collar (to which the guy wires are attached) above

foundation. (fed) r ... moment of inertia of the base of the foundation. (b~sed on

dimensions in feet) j- ratio of lner arm of resisting couple to det>lh dt. (Sec

table 2) lc - factor by which the soil loadi111 due to dead load m.usl be

multiplied to equal tbe soil loading due to overturn•n.r, as follows:

JcS,- S., also (S7)

(68)

Valuea of k for variou• foundation shapes a1·e as follows: 8.15e

Octaron: k- --d- (72)

8.32e IIe:ucon : k = -(1 - (72a)

Square: k = ..!:!~ (72b) d

Circle: k = S.Ole (72c) d

L - lner arm of wind load (feet) to be calculated as follows:

II L~hr+-2-

Lo -perimeter of foundation pedestal. (feet)

'M• ~ bending moment in base. (inch-pounch)

(9)

Mo - negative bending moment at collar. (foot-pounds) (see equa· tion 53)

Mt- overturning moment about base of foundation. (foot·J)Oundo) Mr = Pw L (7J

Mo - maximum positive bendinr moment between collar and base. (foot pounds) (see equation H)

M • ~ bending moment about bast of tower. (fooloj>OUtl<h)

Mt = Pw ( ~) (26)

liLa - bending mom.:nt in base, due to uplift forus. (inch-pounds) N = number of foundation bolu n"" (Eo/Eo) - ratio, n.;~dulut of eluticity of ateel to that of

concrete. P = safe load which each pile will support, (pounds)

20

P .. - total wind load (pounda) to be calculated aa follows : Pw- P• Do H (8)

p ,.. wind prusure on a flat surface. (pounds per oq. lt.) Pc - wind pressure on projected uea of a cylindrical tower.

(pounds per sq. ft.) P•- penetration or sinking of pile under the last hammer blow, on sound wood. (inclies)

P• = ratio, cflective area of reinforcin• steel to effective area of concrete.

P• = ( A.of ba dr)

Ro - hori .. ontal wind reaction or 1hur at base of tower. (pounds) Ro- horizontal wind ruction at collar. (pounds) R• = pull on JUY wire due to wind pressure. (pounds)

R Re • - Sin I or, (40)

R 1 =esc I (41) R.- vertical component of pull on rur wire. (pounds)

R.- ( Rr + Ro) cos I (4S) Rt - initial tension on guy wire. (pounds) r- radius of gyration of the base of tlte foundation (feet). Its

relation to the moment of inertia can be expre&led u followa:

I ~ ar1

rear ran;inr:

r-~! rw ~ width of tower base rinJ. (Inches) S = total unit soil loadinr. (pounds/sq. ft.)

S - s, + Ss also, S -St (I+ lc)

s, =unit soil loAdinr due to dead load. (pounds/sq. ft. )

(1 7)

(25)

(I)

(59)

s,,. ~ unit soil loading due to minimum dead load (r>ounds/sq. h.) to inclu~e the weirht of the empty tower, the foundation and the ~anh fill only. It does not include insulation, platforms, piring, liquid, uc.

So .,. unit soil loadinr due to overturning moment . (poundJ/tq. ft.) So= total maximum unit shurin&" load. (pounds per lineal foot

of ped~ stal peri meter) So - S. + S. (II)

So ~unit shearin&' load due to dead load. (pound• per lineal foot o I ped est a I perimeter)

S w. + w. + w,- <a, so~•• ' ... J.,p (82)

So- maximum unit sheao·ing load due to onrturnio.r moment. (pounds/linn\ foot of Pedestal perimeter). For pedestals of various ohapes, the value• of (S.) are as follows:

Octaron ;

Ilexason:

Square'

Cucle:

Sou = maximum allowable unit soil loadin&'. (pounds/oq. ft.) s~ - load on each foundation bolt. (pounds)

(83b)

(83c)

S, - unit soil loadi"i due to pull on guy wires. (pounds/aq. ft.) s.,.- total unit soil loadift&' under minimum dead load couidtioou.

(pounds/sq. ft.) So ""'unit ttreu in tower shell due to bendin~r moment (Yt).

(poundo/oq. ft.) t = •h~JI thickness. (feet) u ~ bond atress (between concrete and reinforcln~r steel) per

unit of ao·ea of •urface of bar. (pounds) V - velocity of wind. (miles per hour)

V • = vertical shear load outside the criti~al xction (ne Figures 3" and Jc:). (pound5)

W = total .weia-ht on soil (poundo) calculated by' the followin~r equation;

w-W•+W• (3) W • - weight of auxiliary material and equipment supported by the

foundation (pounds), includin• liquid in the tower, insula· lion, platforms, piptng, etc. (Does not include weight of tower.)

W• - we!ll'ht oC hammer. <voundl!) w, = weiaht of foundation pedestal (top course). (pounds) W, =weight of empty tower. (pounds) W • ~ n1inimum dead load .(pou11ds) which is the wei Jilt of the

empty tower plus the weirllt of the foundation, including the earth fill on top of the base.

Z- uction modulus o{ the bate of the foundation (to b~ based on foundltion dimensions in feet) as follows :

z - -1 (It) c rr~ 3.1416 I ~ angle which ruy wil'e maku with the vertical. ( derrees)

l:o- sum of perimeters of noinforcin&' bars withi11 the limits of the beam width (ab). (inchea).

Simplified Design for Tower Foundations Curves reduce design time for octagonal reinforced concrete tower foundations by quick selection of base size, thickness, re­inforcement area and unit bond stresses

Andrew A. Brown, Union Carbide Chemicals Co. South Charleston, W. Va.

DESIGNERS OF FOUNDATIONS have used many different locations for sections and beam widths to compute bond shear, bending moments, and diagonal tension shear. Since agreement on these important phases is not com­plete, this presentation uses The American Concrete Institute Building Code Requirements as a guide for reinforced concrete design and the allowable unit stresses therein. The usual assumptions are followed as to the behavior of reinforced concrete and soils.

For simplicity, the derivations of formulas are based on the inscribed circles of the octagonal base. This does not influence the accuracy of the final results. The foun­dation engineer is ever mindful of the fact that a sub­structure design based on inexact soil bearing determina­tions, concrete with variable strength, and loads which can be off 10 percent or more, is not very definite. The application of good judgment coupled with experience is more important than carrying out computations to more significant places than the information and assumptions warrant.

Foundation Size. As the size of the foundation is the first design requirement after the permissible soil bearing has been established, the formulas used for this determi­nation will be derived in that order.

When the resultant of all forces acting on the founda­tion strike the base within the kern, the forces acting on the soil can be represented graphically by a right regular cylinder resting on an ungula of a right regular cylinder. If it is on the edge of the kern the soil reaction forces form an ungula whose base is a circle; when it is outside the kern, the ungula has a base in the form of a circular segment (Figure 1). The volumes of these solids are equal to the total weight supported by the soil, and their moments about the center of the base are equal to the moments of the external forces acting on the foundation about the same place. Then the eccentricity "e" meas­ured from the center of the foundation equals

external moment of all forces (M) Total vertical load (W)

which is equal to the moment of the forces acting on the bottom of the foundation divided by the total forces acting on the base.

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

+t 0

I

0

... o..,

t-- -t--+--+--'k;.::0_--1 ~ ~

~

~"g ::oCJ ...,_ --= ou.

Equivalent Square

ACI 1208

Section tor Bond ond Moment

ACI 1204 (o),l205(c)

FIGURE 1-The resultant of all forces is within the kern.

radius equal to Ys of the diameter of the circular founda­tion) the maximum soil pressure P is equal to the total height of the right circular cylinder and ungula drawn to graphically represent forces acting on the base.

For this condition the maximum soil bearing

P= - 1+-W ( 8e) ?TR2 D

and the minimum soil bearing equals the height of the soil pressure cylinder or

h=~(1-~) ?TR2 D

Let V equal volume of cylinder and ungala which equals W, the total vertical load.

To get the general formula, let the maximum soil bear­ing equal unity and h equal minimum soil bearing, then the total load

+ ?TR2 (1- h} ?TRI (h+ 1)

W = ?TR2h - ----=--2 - 2

For a value of ~ less than Ys, the maximum soil bear-

ing (unity) can be computed in terms W and D. As an example, for

w e/D= .10, P1 =A(1 + .8) and Pmln. =

W .2W TC1-.8) =-x-

21

SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS .

If 1.8 is reduced to unity or P, then 1

P 1 = -- the m n . 9

height of the right cylinder and the height of the ungala becomes%·

Then

r.R2( I ) W=-2- --g+LO

5r.R2 = --orC R~ 9 v

Where Cv is a coefficient which when multiplied by the product of the maximum soil pressure and radius squared will give W, or the total volume of the cylinder

and ungula for this condition. For the values of ~ , the

coefficients Cv ~ere computed, Column 8 Table I , and C values, Column 9, were obtained by dividing

C, R 2 by 4 ( ~ ) 2•

TABLE 1-coeftlclents for Various e / d Values

I 2 3 4 5 6 7 8 9 ------------ ------v M e/D c v I( " CvR2 C.,Ra M/2V V/ 4(o/ d) •/D CvR• V/ 41e/ d)2 --1------- ---.10 u· 62' .0660 .0603 .467 . 079 .12 1.602 • 27.82 .15 45° 34' .1198 .1045 .436 .167 .II 1.671 34.52 .20 sa• 08' .1823 .1516 .416 .264 .10 1.745 43."3 .25 60. .2618 .1088 .396 .403 .09 1.826 56.34 .30 666 25' .3269 .2448 .374 .583 .08 1.916 74.84 .35 726 32' .4068 .2879 .361 .811 .07 2.013 102.7 .40 786 28' .4904 .3276 .333 1.099 .06 2.121 147.3 .45 84° 16' .5773 .3626 .314 1.463 .05 2.244 224.4 .50 900 .6666 .3927 .291 1.921 .04 2.380 371.8 .65 966 44' .7679 .4172 .276 2.605 .036 2.464 600.8 .60 101• 32' .8604 .4354 .256 3.235 .03 2.533 i03.i .65 1076 28' .9440 .4493 .238 4.170 1-----.70 tta• 35' 1.038 . .4667 .220 6.370 Rcoultant inside korn .75 120° 1.128 .4557 .202 6.97 .80 1266 52' 1.226 .4537 .165 8.94 .85 1346 26' 1.312 .4408 .168 11.65 .90 u3• 08' 1.404 .4297 .153 14.99 .95 154" 09' 1.488 .4107 .138 19.60

1.00 1800 1.671 .3927 .126 26.13 -- ·- . - --Re~ultant outside or on edge of kern

.450

.425

400

.375

.350

.325

.300 •to

.275

.250

.225

.200

.175

. 150

.125

Resultant Outside the Kern. For the development of the equations for moments and total forces acting on the base when the resultant force is outside the kern, refer to Figures 1 and 2, which show this condition.

To get the volume of the ungula of height P, whose base is bounded by the angle ± a as measured from the X axis, we have dV = d A P'. dV is a volume whose area of base is dA and height P' and is located a distance Reo a cp from the Y-Y axis. Then by similar triangles

P' P 1 (Cos</> Cos a)

_.:....:_ ____ .:...; dA=2R Sin</> dx and dx=R Sin</>d</>. (I -Cos a)

p' :

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

c

.100 .10 ~ ro

c 14.75

FIGURE 3-Curve used to determine soil bearing or diameter of foundation base.

22

t

.65

60

.55

50

.45

.40

. 35

.30

.25

.20

.175

. 15

. 125

., 0

7S

.70

.65

.60

.55

.50

.45

.40

.35

k

. . 30

.25

.125

FIGURE 4-Curve plot of Columns 1 and 5, Table 1.

Jy substitution,

2R2P dV = 1 (Cos</> - Cos a )Sin2 </>d</>

(1 - Cosa)

and V = 1 (Cos</>-Cosa) Sin2</>d</> 2R2P )a

( 1 - Cosa) 0

= 2R2P 1 [Sins</> -Cos a ( !_-_I_ Sin</> Cos</> )] " ( I -Cos a ) 3 2 2 0

_ 2R2P1 [Sin Sa Sin a Cos2 -a- a Cos a J - · + -----::----(1 - Cosa) 3 2

Or V (W) = CR2P1 where C =

2 [ Sin3a +Sin a Cos2 a - a Cos a] (1-Cosa) 3 · 2 ... (1)

T he moment of any ungula which represents the forces applil'd on the base of the foundation about the Y-Y axis is the summation of the product of the differential volumes, dV and R Cos q,.

2R3P1 So dM = [(Cos</>- Cos a ) Cos</> Sin2 </>d</>] ( 1 - Cosa)

2RSP } "' and M ( 1 (Cos2 </> Sin2 </>-Cos a Cos¢ SinZ </> )d¢ 1-Cos a)

0

2R8P 1 [- ..:_ ( -1 Sin 4-cf> - cf>) _ Cos a Sin~ "']It

( 1-Cosa) 8 4 3 0

2R3P1

( I- Cos a)

[ a + Cos a Sin a 8- 2Cos3 a Sin a _ Cos a :inS a J

k' O 10 a.. a:

~

I

ACII205~a) 4 5° far Shear .... 2 ACI 1205 (b) .c (/)

.... 0 -".a

• Equivalent Square

ACI 1208

FIGURE 5-The soil reaction is the sum of the forces in the shaded area .

or M = CRSP 1 where C = 2 (1 - Cosa)

[a + Cos a Sin a - 2Coss a Sin a _ Cos a SinS a]

8 3 ... (2)

By use of equations 1 and 2, Columns 3, 4, 5 and 6 of Table 1 were computed for values of K or angle alpha.

M e V Column 5 = 2V = 0 andColumn6 =

4 ( ~ ) 2 •

The curves on Figure 3 were plotted by using Columns 5, 6, 7 and 9. From these curves the size of the founda­tion can be obtained for a permissible soil bearing or conversely, the soil bearing can be computed for a known foundation. The formula for soil bearing is

w Pl = Ce2

where W = weight of foundation and equipment and e = eccentricity caused by wind moment, seismic

forces, etc. and C = a numerical coefficient for the respective e/ D

value. With the maximum soil bea1;ng given

w C=-

Plc2

e e which locates 0 = C and D = c ·

e The relation between K and D is shown by the curve

on Figure 4 which was formed by using Columns 1 and 5 of Table 1.

Foundation Thickness. After getting the SIZe of the hasP., the next step is to determine its thickness. Since the missiblc maximum unit shear is 75 psi this is usu-

23

SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS • • •

(!) Widl~ of fooling Ruisting Shttt • C 1 R (Stt Ttblt 2 Col13l

0 Voluont of Slrtll Prls• Whose But il • Tropuoid ond Segment of Circle • CPsR2(su Tobie

(!) Voluou of Slrtu Pris• Wedge +Seclioo of Unouto •

c ~~ R2 (Set Ttblt 2 Cot. s a 6 I.

2 Col.3).

.40

.35

.30

.25 .25

.l0···-··20 .15 .15

.to .1s .20 .2e .30 .35 .40 .45.50.s6Sl .70 .80.90 to 1.5 25 3.0 3.5 4.04.550 tlrul~tf::E ·10

c

FIGURE ~urves for computing shear for diagonal tension.

ally the controlling factor. In many designs this limits the strength of the concrete to 2,500 psi for the most P.eonomical foundation. The soil reaction considered in computing diagonal tension is the sum of the forces acting between 90 degree radial lines drawn from the center of the base through the two corners of the equivalent square and bounded by section B-B. This section is parallel to the side of the square at a distance "d," (depth of con­crete) from it. One can see by Figure 5 that these forces can be represented by a section of an ungula whose height is P ,, a wedge whose base is a trapezoid abed and

h. h (.707-Cosa)P,

eJg t of ( 1 _Cos«) and a force solid whose area

of base is the area of the trapezoid abed and the segment of the circle whose chord is cd and of height P5 • This

latterareais [ : -(1-2K') 2JR2• The force is R2P6

[ : -( 1 - 2K')2 J and is Column 3 of Table 2. (See

curve 2, Figure 6.) When K is .1465 or less, this area is a segemnt of a

circle. The volume of the section of the ungula can be solved by application of limits of 45 degrees for ¢ in computing the volume.

2R' P C5

V= ( 1 _ Co: a) ) 0

(Sin2 t/1 Cos t/1- Cos a Sin2 t/l)dt/1

By integrating and substituting the values of the trigo­nometric functions for the45 degree angle, the force V =

2R'P ( 1 _cO:«) [.11785- .1427 Cos«]

This formula provided the value in Column 4 Table 2 for the various values of K (a).

The volume of the wedge is =

R2P4 (.707 -Cosa)2(2 Cos .a+ 2.828) 6(1-Cosa)

and for the respective K and a values the volumes are recorded under Column 5, Table 2.

These two columns are added (See Col. 6) and the results are plotted producing Curve 3 of Figure 6.

The width of the footing "b" for diagonal tension is 2R ( 1 - 2k). When a is 45 degrees or less, it is 2R Sin a. These values form Curve 1 of Figure 6 and are tabulated in Table 2, Column 13.

Bond-lending Moment. The slab is now investigated to determine the area of reinforcement and unit bond stresses. The moment of all forces to the right of Section A-A, Figure 1, determines the area of steel, and the sum of these forces is the shear used in computing the bond. Section A-A is located by passing a vertical plane through the foundation along the side of the equivalent square. The external forces acting on the base can be conceived

TABLE 2-Values to Calculate Diagonal Tension, lond, Moment and Beam width

1 l 3 4 5 6 '1 8 9 10 1l l l 13

Col . Mom. v Sec. Unit. weclr 4+5 Ungula ~.Cy1. Un l M. lRaln• Octaton .. b ..

K • CP•R' C'P•R' c•p, • CP•R• CPoR• PaR1 CPt • SeQ. Cyl. CR CR CR

.10 a&• 52' .1635 .0660 ..... .0660 .0668 .1635 .0076 .0132 1.20 1.28 1.20

.15 411° 84' .2954 .1198 :oio3 .1198 .1198 .29115 .0203 .0361 1.40 1.43 1.40

.20 63° 08' .4264 .1611 .1804 .1823 .4473 .0422 .0729 1.60 1.63 1.20 .24 60" .53M .1860 .0547 .2407 .2518 .6142 .0729 .12119 1.73 1.83 1.00 .30 66° 25' .62M .2029 .0951 .2980 .3269 .79?:1 .1140 .1961 1.83 2.00 .80 .35 7-;;- 32 .6954 .2144 .1353 .3497 .4068 .9780 .1658 .2829 1.91 2.00 .60 .40 78" 28 .7(54 .2232 .1726 .3958 .4904 1.173 .2294 .3924 1.96 2.00 .40 .45 84° 16' .17M .2302 .2067 .4369 .5773 1.389 .30t9 .5195 1.99 2.00 .20 .50 90" .7854 .2~7 .2~7 .47M .6666 1.671 .39?:1 .6666 2.00 2.00 .00

Dlneooal Tension Bond Bendin& Mom. Width Beam

24

®volume of Strtn Priam Whou Bou it Segment of

Circle , uud in Computing Shear • C p3 R 2'

(!)Volume of Siren Prism Which is on Ungula , ustd CPgR2 in Computing Shear •

@ Width of Footing ot Section A-A used lor Shear and Bonding Moment • CR

c

(!) Moment of Sinn Priam (Ungula) About Section _ . _ CP3 R3

A A , Fog. I - IO

(!) Moment of Strtu Prism , Section of Crlindtr CP R3

About Section A-A • ~

.50

.45

.40

.35

.30 k.

.25

.20

.15

.10

FIGURE 7-Curves for computing shear for bond and bending moment for reinforcement.

as being in the shape of an ungula of height P 2 and a segment of a right circular cylinder of height P3• The sum of the two volumes is the shear force, and the sum of their moments about A-A is the bending moment that determines the reinforcement. The weights of the con­crete slab and earthen fill are deducted from the vertical forces. This is easily accomplished by reducing the inten­sity of the uniform bearing load acting ~n the bottom of the base.

The volumes for the ungula have been computed earlier for obtaining soil bearing and those values for K equal to or less than .5 are shown in Column 7 of Table 2 and Curve B of Figure 7. {See Equation 1 for V.)

The volume of the segment of the cylinder is equal to the product of the area of the segment and P8• The area is easily computed by making use of the fact that the middle ordinate is KD. Values for the respective K" are shown in Column 8, Table 2 and plotted as Curve A, Figure 7.

The bending moment equation is derived by substitut­ing R (Cos 4> - Cos a) for R Cos 4> in the development of the formula for moment about the center of the foun­dation.

h. . 2R3Pl ~a • d T IS giVes M = (Cos¢ - Cos a) 2Sm2 ¢ ¢ (1- Cos a) 0

_ 2RBP1 [ 1 ( 1 . ) 2 Cos a Sins¢ - (1-Cosa) - 8 T 5m 4¢-¢ - 3

- Cos2 a ( : - + Sin 2¢) J:

M= 2R

8P1 [.::_(1+4Cos2a) -2-SinaCos3 a-

(l-Cosa) 8 4

- osa msa 2 C s· + Sin a Cos a] 3 8

By substituting in this equation the trigonometric val­ues for the respective angles corresponding to the K", Column 9, was obtained and CurveD plotted on Figure 7.

The moment on the forces whose configuration is a segment of a cylinder (see Figures 1 and 2) is derived as follows:

dM = 2RSP8 ~~Cos¢- Cos a) Sin2 ¢ d ¢

[SinS¢ - Cos a(¢ --1- Sin 2 ¢) ]a

3 4 0

[ Sin

3s a + Sin a Cos2a

2- a Cos a J = 2R3P3

The values obtained for the angles a (K) are noted in Column 10 and form Curve E, Figure 7.

The widths of the foundation at the sections are equal to 2R Sin a and are shown in Column 11, Table 2, and Curve C, Figure 7. Column 12 and the dotted curve (Figure 8) indicate the width of beams for any octagon.

Use of Curves. As an illustration of application of the curves, the following information is given: height of ves­sel, 112 feet; diameter, 8 feet; the anchor bolt circle requires a 10-foot octagonal pier; top of pier is one foot above grade and 6 feet, 6 inches above the bottom of the foundation; permissible soil bearing 3,000 psf (P1 ) at 5 feet, 6 inches below grade; wind pressure, 30 psf of hori­zontal projection of the vessel.

Operating weight of vessel, 200 kips; vessel empty, 100 kips; and test weight, 300 kips.

The diameter of the base required under operating conditions will be determined first.

The moment of wind force about the bottom of the foundation is 112 x .03 x 8 x 62.5 = 1680 i·ap-feet.

25

SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS .. •

Estimate the weight of the foundation using a 22 foot, 6 inch octagon, two feet thick.

Pier= (82.8) (6.5) (.15) Slab= (419-82.8) (.3) Fill = 336.2 ( .35)

= 81 kips = 101 = 118

Total = 300 Weight of vessel (operating) = 200

w =500

E . . 1680

ccentnCity e =-- = 3.36 feet c2 = 11 3 500 ' .

c = 500,000 (11.3) ~,OOO = 14.75. With this value of C, ejD

is obtained from the curve on Figure 3 as .153. Then 3.36

D = - = 21.9ft . . 153

Next try a 22 foot, 0 inch octagon with a thickness of 1 foot, 6 inches.

The weight of concrete and fill becomes 280 kips and W = 480 kips. To compute the maximum unit bearing

1680 e 3.5 e = 480 = 3.5 feet 0 = 22 = .159.

From the curve used above, C = 13.5 and 480,000 . . .

P,= 13

_5(3.5)2 = 2,900psf<3,000. Thts IS cons1dered to

be near enough to the allowable soil bearing. To strive for closer agreement is believed to be inconsistent with the accuracy of the established bearing value of the soils and therefore would be a waste of time.

The unit bearing 1.mder the foundation for test condi­tions and one-half of maximum wind load is found to be 2,370 psf.

To investigate the 1 foot, 6 inch slab for diagonal tension, the area of the 10-foot octagonal pier is used to compute the side of the equivalent square of 9.1 feet. With the e/D of .159, K is found to be .88, by use of Figure 4 and KD = 19.35 feet. The distance from the

About the Author Andrew A. Brown, Captain, Civil Engineer Corps,

United States Naval Reserve, is Public Works Offi­cer, Naval Training Center, South Charleston, W. Va. and an engineer in Design and Construction, Union Carbide Chemicals Company, South Charleston. Mr. Brown's professional experience includes several years in the Bridge Department, State Road Commission of West Virginia. He has been a consultant on bridges for several cities. During his 12 years of active duty in the United States Navy some of his billets were: De­sign and Construction Officer, Fifth Naval District, RO in CC, Naval Brown Missile Test Center, Point Mugu, California, and Public Works Officer, Naval Station, San Juan, Puerto Rico, and Naval Air Station, Kaneohe Bay, Hawaii. He is a member of International Association for Bridge and Structural Engineers and has BSCE and CE degrees from West Virginia University.

26

center of equipment to the point where the diagonal 9.1

tension is computed is-2-+ 1.17 or 5.72. Then K'D =

11.0-5.72 = 5.28 and P. = -- 2,900 = 790 psf, ( 5.28 ) 19.35 .

and P5 = 2,900 - (790 + 625) = 1,475 psf. K' = 5.28 22.0 = ·24·

By referring to the curves for computing shear for diagonal tension (Figure 6) and using K = .24, the width of the footing resisting shear is 1.04 ( 11.0) = 11.45 feet. The shearing force is= ( .515 ) (1,475) (1P) + (.23) (790) ( 1 F) = 92,000 + 22,000 = 114,000

114,000 . . pounds. V = ( 11.45) (14) (12 ) (.88) = 68 pst < 75 pst

maximum allowed. The section for computing bond and reinforcement is

taken along the side of the equivalent square, A-A Figure

( 9.1 ) 1. Then K'D = 11.0- -2- = 6.45 and K' = .293

( 6.45 ) P2 = 19.35 2,900 = 965 psf

P3 = 2,900- (965 + 625 ) = 1,310 psf

By use of the curves on Figure 7, the shear for bond, the bending moment and width of beam are computed for K' = .293.

Width of beam = ( 1.82 ) ( 11) = 20.0 ft. (circle): ( 1.96) ( 11) 0 = 21.55 ft. (octagon )

(3.15) (965) (112) Shear for bond = (.76) (1,310) ( 112) + '---'---~--'--

10 = 120,600 + 36,800 = 157,400 lbs.

'8 d' (1.08) (965) (113) en mg moment -

1 O

+ (1.83) (1,310) (1 JS) 10

= 139,000 + 319,000 = 458,000 ft.-lbs.

458 Area of steel required per foot =

(20) (14) (1.44) = 1.14 sq. in. per ft. of width.

Since wind forces contribute more than 25 percent of the moment, stresses can be increased one-third so the area becomes (. 75) ( 1.14) = .85 sq. in. A six-inch spac­ing each way of No. 6 bars = .88 sq. in. ~0 = 4. 7 inchcE

. 157,400 and bond stress 1s = (20) ( l 4) (.88) (4.7) 137 psi.

Some foundation engineers prefer to base the steel and bond requirements on the middle one foot wide strip. Under this condition the force for bond is = ( 1,310) (6.45) -J (Y2) (6.45) (965 ) = 8,450 + 3,110 = 11,560 pounds.

The bending moment is = (6.452) (Y2) (1,310) + ( Y3) (6.452

) (965 ) = 27,300 + 13,400 = 40,700 ft.-lbs.

(.75) (40.7) . . . As = ( 14) ( 1.44 ) = 1.51 sq. m., a five-mch spacmg each

way of number 7 bars = 1.44 sq. in., ~0 = 6.6 inches.

11,560 . u = (6.6) (.88) (14) = 142 pst.

The design for the top of the slab reinforcement, "top bars," which arc required by certain combinations of loading, is left for the reader. # #

Calculation Form for Foundation Design

For complete design of octagonal foundations for stacks and towers or for estimates · only, this form will solve the problem easily and quickly

Bernard H. Shield, Celanese Chemical Co., Pampa, Texas

IN THE DESIGN OF FOUNDATIONS and structures for chemical plants, the structural engineer normally is not too concerned with a highly theoretical, or complicated mathematical approach. From a. practical standpoint, the design assumptions quite often are not accurate enough to justify such an approach.

Since the chemical industry is such a fast moving, often changing, and complex field, the design engineer often lacks sufficient time to make an accurate theoretical analysis or sometimes even a very thorough practical analysis. Quite often he must wade through a lengthy article or text concerning an unfamiliar problem, or a problem which he has not worked recently. While the time schedule suffers and other details of the job are neglected, he must set up the problem for practical analysis. For many problems of a repetitive nature, much time is consumed in setting up the sketches and frame­work for an analysis rather than in simply solving the problem.

How many times have you heard the question, "VVhen will the foundation drawings be out?" I have heard it many times, quite often as soon as a request for appro­priation for a new installation is approved. Faced with this situation, the engineer must constantly seek solutions to his problems that will give safe and economical designs and use a minimum of his own time.

The following calculation form for octagonal founda­tions for towers and stacks was devised with this idea in mind.

We have used the prototype of this form quite success­fully for about seven years and believe it is worthwhile to pass on to others. The. form is largely self-explanatory with the nomenclature and design method being explained as the solution progresses.

Design Basis. The following general comments should be of help in using the form the first time. Moments are computed about the centroid of the base of the pad, ignoring any shifting of the neutral axis as loads are applied. Soil stresses are computed using the section modulus of the base pad around its axis of symmetry. The slightly higher soil stress which would be obtained by using the section modulus around a diagonal is ignored. Stresses caused by a moment in the base pad are computed according to the ACI code by computing the moment along a line which would coincide with the side of a square of equivalent area to the pier. Two-way reinforce­ment is then provided similar to the normal method of reinforcement for two-way reinforced footings.

In computations of forces, the area and stress diagrams are divided into simple geometrical shapes for ease in computation. The design of tensile reinforcement in the pier is a practical rather than a theoretical approach. Anchor bolt lengths and hooks are designed according to the ACI code for hooked plain bars. The length will depend upon the design stress used for the bolts, so if a designer wishes to use a stress which differs from that shown on the anchor bolt table, he may easily change the length.

If he desires additional safety, he may choose to use a lower design stress for sizing the bolts and use the lengths given in the table. I prefer using higher anchor bolt stresses than some designers, taking advantage of the V3 increase in allowable stresses for combined loading in which wind is a factor. This will of course give anchor bolts which are smaller in diameter and longer in length.

I have a great deal of confidence in the reliability of de­sign stresses in steel but very little confidence in the al­lowable bond stress for a smooth bar. Many times anchor bolts are installed without proper cleaning and with thread cutting oil all over them. So, who knows what bond stress will be developed?

I believe much work remains to be done to devise, and prove by tests, a really good method of design for large anchor bolts. In the meantime, I prefer to use a design which I believe to be safe and economical, and recognize the right of other engineers to use their own criteria.

It should be noted that the use of this form is not limited to the complete design of a foundation. Should it be desired to obtain only the size of the foundation pad, for such things as estimating or layout, one need only proceed through Step 5. Step 15 with Figure 3 are quite useful to transmit information to a draftsman, and the anchor bolt tables are useful in fabrication of anchor bolts.

The next time you have this type of design problem, give this form a try. It is easily revised for special cases. You may not appreciate its merits so much if you only have one foundation to design. If you have two or more, I think you might begin to like it. If you have 50, you will probably become downright fond of it!

Procedure. Considerable time and effort are usually re­quired to make a detailed and accurate design for octag­onal foundations for towers, tall reactors, pressure vessels, or stacks, particularly if the designer is unfamiliar with the problem. Consequently, a complete design is often not made, and this may lead to either an unsafe or un­economical design or both. This method provides suf­ficient des.ign detail for a safe and economical design. A relatively inexperienced designer can use the form, but such work should always be checked by an experienced designer. This form makes such checking easy. The finished calculation provides a neat, understandable, and legible record and should be maintained for record pur­poses.

This form is intended for the complete design of foun­dations which have relatively large base pads in relation

27

CALCULATION FORM FOR FOUNDATION DESIGN • .•

to pier sizes. It can also be used for foundations which have relatively large pier sizes in relation to pad sizes; however, for this condition the designer must be alert to make necessary changes in the calculations. Referring to Figure 2, the changes which will be necessary are as follows:

When D2 > 0.45 D1, d will be a negative number, and along with area (3) will di-op out of consideration. Di­mension C will also have to be computed by other means than as shown. Therefore the calculations for shear and moment as outlined is Step 7 will have to be altered.

When e is zero or a negative number, the calculation for shear is unnecessary and Step 11 may be omitted.

An orbltrorr 1ft. Is added Dotl• for vapor lines. loddert p lotformt~t_!ltc~.==::::_- !=~;:;:;:~I

..: -• lnsul. ... • .. ..:

! 'a

... .!

... • • 0 ..

Do- Dlo. outelcle lntuL, ft

P..•P4(T4)(q,tl) lb •

FIGURE 1-Record dimensional data on this figure.

Step 1. Record dimensional data on Figure 1. Depth below grade, h5, should be determined by a reliable soil survey for the site. The pier diameter, D2, is usually about 1'-6" larger than the tower skirt diameter. The pad diameter, D 1, and pad thickness, ht, must be assumed and solved for by trial and error.

For selecting an initial trial pad size, the following method is suggested.1

Where wind load is likely to govern:

D1 = Trial dia. across flats, ft.

Mr =Total moment about base (See Step 2), ft.-lbs. S =Allowable soil stress, lbs/ ft.2

(Suggest using 1,800 lbs ft2 for first trial.)

Wind load computations are based on the A.S.A. bul­letin "Minimum De~gn Loads in Buildings And other Structures," A58.1-1955.

28

DIAGONAL TENSION

0 u

FLEXURE 80 0

DIAGONAL TENSIO

e

FIGURE 2-Stresses due to flexure, bond and diagonal ten· sion are computed along these sections.

R efer in the ASA bulletin to Figure 1 and Table 3 and list wind pressures for each height zone for the plant site as follows:

For height zone, T 1,

P1

= 0.6 (• ) = _____ lbs./sq. ft. • Value from Table 3

For height zone, T 2,

P2 = 0.6 ( ) = __ , ___ lbs/sq. ft.

For height zone, T 3 ,

P8 = 0.6 ( ) = _____ lbs/sq. ft .

For height zone, T 4 ,

P4 = 0.6 ( ) = _____ lbs/ sq. ft.

Compute and record on Figure 1, the values of Ph P2, etc., and the values of L1, L2, etc., for the size tower being used.

WR = Weight of reboiler, full of water = lbs.

Sten 2. Compute total overturning moment, Me.

Mw = Moment due to wind:

Lt (Pt) = X ---------------ft.lbs. L2 (P2) = X ------------- ft.lbs. L8 (P8 ) = X ------------- ft.lbs.

L, (P,) = X --------------- ft. lbs.

Tot. Mw

Ma = Moment due to reboiler weight:

Wa ('C) =------X------ -------ft. lbs.

=Total=Mw+MR = ft. lbs.

Step 3. Compute vertical loads.

A1 = Area of base = 0.828 (D12) = 0.828 ( ____ ) =-----·- sq. ft.

A2 = Area of pedestal = 0.828 (l;>l) = 0.828 ( ) = . sq. ft.

= At-A2 - --- - ------- _ sq. ft. A8 = Area of fill

WB ·= Wt. of base =A1 (h1 ) 150= __ X ___ X 150= ___ Ibs.

Wp = Wt.ofpedestal =A2 (h2 ) 150= _ _ _ X ___ X 150 = ___ lbs.

Wp = Wt.offoundation = WB+Wp +--- = lbs.

Wc=Wt.offill =A3 (h3 ) 100=--- X--- X 100=----- lbs.

WT= Wt. of tower at time of mounting on foundation ..... = ______ lbs.

W A= Wt. of tower accessories and contents installed after mounting:

W a = Reboiler wt.

Ww = Hydro. test water

Insulation

Piping

Platform and ladders

Other

WA, Total =

. _____ lbs.

------- lbs.

- - ----- lb$.

lbs.

___ ____ lbs.

_______ lbs.

W = Max.wt. on soil= WF+wc +WT+WA

________ lbs.

, _____ lbs.

Subtract Ww ------lbs.

W8 = Wt. of finished tower, empty

W L = Wt. of liquid when operating

W0 = Wt. of tower, operating conditions= W8 + WL

Step 4. Compute maximum soil loading.

SA= Allowable soil stress at site and depth = ___ lbs./ft.2

Neglect maximum theoretical loading with tower full and full wind load unless tower operates liquid full

- ______ lbs.

=+ ____ lbs.

------- lbs.

Section modulus of fdn . pad= Z = 0.109 (D1') = 0.109 ( __ ) = -----ft.•

Case 1. Tower operating with full wind load.

S1 = Soil stress due to load = W0/A

1 __ __, ___ = ______ lbs./ft.2

S2 = Soil stress due to wind= Mc/Z - ___ ! ___ = --- lbs./ft.t S = Maximum stress = sl + s2 = _ __ ,/ ___ = ______ lbs./ft.2

Case 2. Tower tested full with negligible wind.

S = W/A1

- _______ ! ·------------ lbs./ft.2

Use maximum, Case 1 or Case 2, must be~ SA

29

TABLE 1-Type 1 Bolts with 180° Hook

Projection JThreada ____________ jn, __ ~r~l:=t·

Sleeve 0

1

-dla.x_ln. 1

I

I II LJ

in.

D= 6d ±. 1/211 for bolts 1/2" to 7/8" D= 8d ± 1/2" for bolts 1" to 2 1/2"

Step 5. Check stability. The most unstable period is usually just after mounting the tower on its foundation, prior to adding the weight of accessories.

Erection smln. = (W-w A)/Al -S2 = ( __ - __ ) ! ___ - ___ = ___ lbs./ft. 2

Smto. (Soil stress on windward side) must be~ 0

This computation will give a resultant minimum soil stress which is on the safe side for stability because the vertical load is computed with the tower stripped of accessories and the over­turning moment used in computing S2 includes moment due to

Step 6. Compute dimensions and loads for computa­tions of stresses in foundation pad. Compute size of square pedestal with equi'"alent area of octagonal pedes­tal, A 2• (See Figure 2.)

(ACI-318-PAR 1208 (a))

a="VA; = V ft.2 =---------ft.

b= .414 (D1 ) = .414 (---_______ ft.

c = .707 (b) = .707 <---) ---------ft.

---=-------ft.

= - _______ ft.

m =a+ 2 d2 (d2 is in ft., here)

= - -+-------- ft.

y=b+c

---+---- --------ft.

S.= WaA+lwr =-----------~ lbs./ft.2

S4 = S1 + S2 - S3

= ___ + _______ = ____ lbs./ft.2

30

See "Anchor Bolu Detailing Dimensioru" on Ia$! page of thiJ article

Slze Thd. Net••• Hook Mln. Total• d Serte1 Area A J D Y L•• Len~th ---~· 0.120 o·-1• o•.4• o·-a· 0'·2· o·-1·

"'. 0.202 0'-9• 0'-s• 0'-4' o·-a• 0'-9•

". o.302 0'-10• o'-6' 0'-5' o·-a· 1'-o----~· 0.419 1'-0· 0'-7' 0'~· 0'-4• 1'-2. ,. O.SlH I '-5' 0'-10• o•-a• 0'-4• 1'~·

1~· o.12s 1•-1• o·-11• 0'-9' O'~· t•-s• t}( • 0.929 1'-8' 1'-1· 0'-10' 0'-5' 2'-0'

1"'. ------1---t---11---1---1.155 t•-u• t'-2• o·-u• 0'-6' 2'-4'

~~· 1.405 2'-1' 1'-3• 1'-0' 0'-6' 2'-7'

'". t.980 2'-s• 1'-s• 1'-2• 0'-7• 3'-a•

2' 2.652 2'-Q• l'-8" 1'-4' 0'-8. 3'-10'

2"· 3.42:1 3'-t• 1'-ll' 1'-6. 0'-9" 4'·6·

2~· 4.292 3'-5" 2'- l" 1'-8" 0'-10" 5'-2"

accessories. Should the designer so desire, the moment can be reduced for this check by substituting the tower djameter for (D0 + I) in Figure 1, eliminating rcboilcr moment and recom­puting moment on tower in a stripped condition as it nonnally would be during erection. If this jg done, stability under operat­ing conditions should also be checked as follows: (See Case 1 above)

S1 = lbs./ft.

Subtract S2 = _________ Jbs./ft.

Operating Smin. = lbs./ft. and must be~ 0

--( ___ ) ______ lbsft2

~a+d(S) ~D. •

+ __ ( __ _ ) = ______ Jbs ft2

S7 = Sa - SG =--- ___ Jbs/ft2

S8=S4 - S0 = __ _ ---- ___ lbs/ft2

S ~ 4 + dz(S ) (d . . f 9 = !/: D • ~ IS m t.)

2 1

+ - --- _ ( ____ ) = _____ Jbs ft2

+ + - ------- ___ ) = ____ Lbs/ft2

-------_____ lbs/ft2

- ----- ______ lbs/ft2

TABLE 2-Type 2 Bolts with 90° Hook

Projection ------~n.

Sleeve 0 -dia.x_in.

D= 7d ± 1/2"

_____ in.

I /

J

y 12d min.

Step 7. Compute total moment in pad along a section at the edge of the equivalent square pedestal: See Figure 2, Section R-S.

S ize T h d. d SHiea

Net••• Area -0.126 --0.202 --0.302 1-

0 .• 19

1' t--

O.Ml ---1}i' 0.728 --

0.929

l.lM --1.405 --

--- .~~ 1~·

~~· Jj~

~~· ~;;; ·-< --- "-~

1.980 --2.652 --3.423

~~· u~ 2" gf-oj 23(' UooU

-2~· 4.292

See "Anchor Bolu Detailing Dimensions" on last page of this article

Hook Min. Total• A J D y L .. Lenath ------------0'-9' 0'-8' 0'-4' 0'-6' 0'-7' --- - ---- - -0'-11• 0'-11' o·-~· o•-a· o-:9• --- - - - ------1'-1' 1'-0' 0'-5' 0'-9' 1'-0" --- - ----!-- -1 '-3· 1'-2' 0'-6' 0'-11' 1'-2' --- - --I--- - -1'.&· 1'-~· 0'-7' 1'-0" 1'-~·

1'-8' 1'-6' 0'-8' 1'-2' 1 '-s· 1'-10' 1'-9' 0'-9' 1'-3' 2'-0'

2'-0' 1'-11' 0'-10' 1'-5' 2'-•· --------1---2'-2" 2'-1" 0'-11' 11-6 1 2'-7' --------1----2'-7" 2'-5" 1'-0· 1'-9' 3'-3' --- - ----- - --2'-lt" 2'-9" 1'-2" 2'-0" 3'-10" - - --- - ------3'·4" 3'-lA' 1'-4"' 2'-3" 4'-6" ----- - - - ----3'-8' 3'·6" l'-6. 2'·6· 5'-2'

Computations are made according to "Building Code Requirements For Reinforced Concrete," ACI 318-56.

AREA IN SQ. FT. X LOAD,Ibs. = (SHEAR, lbs.) X ARM, ft = MOMENT, ft. lbs.

( 1) = (c X b) X 56

- -- _x ____ x -( 1) = (c X b) X l4S8

- .X X -(2 ) = (c )2 X 56

= (_ __ )2 X -(2) = (c)2 X Y3S8

= ( ___ )2X

(3) = (d X D1 ) X S;;

= ___ X X

(3) = (d X D1 ) X l4S7

= ___ x X -Totals V=

Step 8. Compute depth of pad required for flexure.

d2 req'd = /~ '\J KD1

d2 req'd =~-~x-=--=v

in.

ftc = 28 day compressive strength of concrete = _ __ lbs/in2

fs = Allowable tensile stress in steel. - ____ lbs/in2

D1 is in feet in formula, see Fig. 1; MB is in ft. lbs.

The following factors may be obtained from tables in Concrete Design Handbooks or computrd as follows:

X ( l4c + d)

X ( __ + __ )=

X ( 7'!Jc+ d)

X ( ___ + _ _ )=

X ( Ysc + d)

X(_ .+ __ )=

X {Y2C -t d)

X(_ . + _ _ )-

X l4d

X _

X 31d

X -lbs. Mu = ft. lbs.

K = fc/ 2 kj = _ --..J 2 X _____ X ___ _

- ____ lbs/ in2

fc = 0.45 ftc = 0.45 X _ __ _

k

-------

--~- - ---------+

{$ I ­

nfc t + ______ _

j = 1 - Ysk = 1

- ----

d 2 + bar diameter + 3 in. must be = or < assumed h 1

____ + + 3 in. = in. Actual d2

used

·--- ----- in.

31

CALCUlATION FORM FOR FOUNDATION DESIGN •.. Step 10. Compute bond stress for steel design above:

:EO = Sum of bar perimeters. Step 9. Compute steel required for bottom of pad for

flexure. (Ref. ACI 318, par 1204-e) =Number of ban across D1 X bar perimeter.

_ 12 M8 _ 12 ( ) A - 0.85 c-( ') - 0.85 --'7--:---:-:---'-

• J d2 _ x _ x_ ----- -X. ---- - - -----in.

----in.2 p. = Computed bond stress, lbs./in.z

v M8 is in ft. lbs. See Step 7 • = l;Ojd2 --- X --- X - -

d 2 is in inches. Sec Figure I. - ______ lbs./in.:

Use ___ No. ___ ban at _ _ _ in. o.c. (distribute

uniformly) both ways across D1

•

*Allowable p.----·---- lbs./in.2

V =Total Shear, lbs. See Step 7.

d2 is in inches, See Figure 1.

32

Steel supplied =··---- X -----_ ___ in.2 *Ref. ACI 318, Sec. 305

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

Area on plan.

(4) + (5)

Stress indicated on stress diagram.

(The numbers here are for defining the geometrical shapes on Figure 2. Fill in

Volume of area-stress. Geometrical shape

m+y =S9-- (e)

2

Shear for diagonal tension

blanks at right for computation.) =-2-(_ + __ )(_) = lbs

s (4) Su =~(e)(m) 2

=-2- ( __ ) ( __ l= - lbs

(5) Su = 2 S11 (e)2

3

=~ (_ H= lbs

(6) + (7) s1o = s1o ( b-; Y) ( T) =-4-(_+ __ )(_ )= lbs

(6) sl2 = s~2 ( T) (b)

= -4-( __ ) ( __ )= ___ lbs

( 7) s =....!!. (c)2

12

--12- (_

V1 = Total shear for diagonal tension= Total of above---··· Compute diagonal tension shear unit stress, v:

)'= ___ lbs

- ___ lbs

v V1 =Total shear across Section f·g (See above), lbs

m = Length of Section f.g (See FIG. 2) = _____ inches v=--•-mj d2

d2 is in inches, See FIG. 1.

-----:-------- = _ __ lbs/in.,2 ___ ) ( ___ ) ( ___ ) v =

Allowable v = _ ___ lbs/in2

Ref. ACI-318, Sec. 305.

Step 12. Compute ftexure in top of pad due to uplift. W'=--- ( 150) +--- (100) = ___ lbs/ft2

Figure 2, Section R-S. W' =Uniform downward load on areas ( 1), (2) and (3),

Figure 2.

= h1 (150) + h3 (100)

See Figure 1, and using an average weight of reinforced concrete = 150 lbsjcu. ft. and an average weight for earth fill of 100 lbsjcu. ft.

AREA IN SQ. FT. X LOAD, lba ft2 X ARM, ft. = MOMENT, ft. 11>4.

(1) bXc XW'

-·-X __ X ___ _ x(-2 +-)=--(2) XW'

____ )2 x ____ x(-3-+--)= ------(3) XW' X d/2

__ X __ X ____ X---2----

Mu, Total moment - ------- ft. lbs.

d 1 is in inches. See Figure 1. Step 13. Compute steel required for flexure in top of pad. Use No. bars at ____ inches on center.

12 Mu 12 X A's = 0.85 fs j d

1

= 0.85 -..,..x..,----..,...X..,--- - ____ in.2 (Distribute uniformly across D 1 both ways)

Steel supplied = Number of ban X Area each bar.

Mu = Moment due to uplift across Section R-S, Figure 1. - ____ x ____ = _____ in.2

Step 14. Compute size of anchor bolts.

Refer to Figure 1 and compute moment at base of tower as follows:

M8 w =Moment at base due to wind:

(Ll - h, ) PI = <---­

(L2-h,) P2 = (--­

(L3-h4) Pa=(----

(L,-·h,) P4 = (--- ----

------- ft. lbs.

ft,,lbs.

_______ ft . lbs.

ft. lbs.

Mnw = _______ ft. lbs.

M 8 = Moment due to reboiler from Step 2 - - - - - - • = ft. lbs.

MT = Total moment at base= Mnw + Mn - - - - - - - = ft. lbs.

N =Number of bolts = ("N" should never be less then 8 and preferably 12 or more)

Db= Diameter of bolt circle - ___ _ ft.

Fb =Tensile force, due to MT, per bolt= * (~ MT/NDb) ­(W-r/N)

= (4X __ ; __ X __ J-( __ ; __ )= _ ____ Ibs.

FT =initial tensile force due to tightening nut. - - - - - - - = (5,000 lbs. is suggested for "FT")

____ lbs.

F=Total maximum tensile force= Fb + F-r- --- - - -----­Net Area (At root of thread)= F/Allowable stress

lbs.

• 2

- -------~------- - ----m.

Size of anchor bolt (Add }ln to size determined above for corrosion.)

- -------in. dia.

Net Area of selected bolt. = - --------in.2

See Tables 1 and 2 for detailing dimensions of anchor bolts, and for net areas to use in selecting bolt sizes. • Thi• simplified formula is not exact, but is always on the safe side.

33

.. ::t 0 .. CIO c!

Uae _____ "dia. •-----" long galvanized Iron sleeves.

,....-Fin. EleY. = No. Dio.

..: 1----~-==*r- uae __ ---"dia. bolts

1------rt-....ott-- .#3 ties at 12" o .c .

-if-~;:::;:::::;::::::;::::::~::Dt:==~~jjb1~-;;_7T~;;~;;.; " o .c. both way 1.

ELEVATION

Step 15. Pe~estal steel and reinforcing steel placement.

For vertical steel in pedestal use greater of following two steel areas:

( 1) A8 =Net area of anchor bolts.

--------- in.~

(2) A8 =No.5 bars at (max.) 6 inch spacings.

---------in,2

Use No. bars distributed uni-formly around pedestal in octagon as shown (Figure 3)

As = , _ ____ in.2

Anchor Bolts Detailing dimensions. Tables 1 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 area = an, in.2

** Design basis for L: = Computations are made using an allowable stress of 26,600 psi. Allow 10,000 psi for the hook

About the Author Bernard H. Shield is an engineering group leader

with Celanese Chemical Company, Pampa, Texas. He supervises a departmental group handling mechanical design phases of plant alterations and expansions including project engineering, me­chanical, electrical, and instrumen­tation. Mr. Shield holds a B.S. de­gree in civil engineering from The University of Texas. He worked with the Surface Water Branch of the USGS and instructed in the Civil Engineering Department of the Uni­versity 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.

34

Shield

----" O .C. both WO)'I.

Thll 1ketch is for u .. in drafting finished drawings.

i

PLAN

FIGURE 3-Foundation details.

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

L = fs an - fsliag 7rdp

Where: f8 =Allowable stress at root of thread= 26,600 psi. a.,= Net area at root of thread, in.2

p =Allowable bond stress for 3,000 psi concrete= 135 psi.

f8li =Allowable stress over gross area due to hook = 10,000 psi.

ag =Gross area of bolt, in.2

ACKNOWLEDGMENT There haw: been many fine articles published on this subject and 1 wish tc

acknowledge the information which I have gained from them. In addition, good comments from Celanese engineers, particularly the late Eddie Bayers of Charlotte, North Carolina, and Willard Johnson and Womac Soward with Celanese at Bishop, Texas, have contributed to the development of this paper. I abo wish to recognize the work of our draftsman, DOn Stafford, in the drafting of this form.

LITERATURE CITED 1 Wilbur, W. E., "Foundations (or Vertical Vessels," PETROLEUM RutNI!.R,

34, No. 6, 127 (1955).

Use Graph to Size Tower Footings

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

J. Buchanan, Newcastle University College, Newcastle, N.S.W., Australia

GRAPHS OF SIMPLE dimensionless numbers may be used to size spread footings. These numbers describe the action of the footing under a known load system and allow the user to select a footing size that will maintain stability without exceeding a specified maximum allowable soil bearing pressure.

A typical footing arrangement is shown in Figure 1. As in the usual treatment the soil under the footing is taken to be perfectly elastic, and no credit is allowed for the soil lateral support. When the moment (M) is negli-

FIGURE 1-Typical footing arrangement for tall towers.

gible compared with the dead weight (W), the soil bear­ing pressure is uniform over the whole area of the base of the footing as shown in Figure 2(a). As the moment increases, the soil pressure distribution changes, as shown in Figure 2(b), (c) and (d), until it reaches the extreme (and in practice impossible) case shown in Figure 2 (e) where the structure is just balanced on one corner of the footing and the :.,caring pressure is infinite at that point.

If M is increased further the structure must topple. So long as the allowable pressure is not exceeded, all

of these possible arrangements are inherently stable. How­ever, at some stage in the sequence, the maximum soil pressure, that is the pressure at 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.

For any particular footing shape (square, octagonal, etc.) and orientation, a pressure pattern as shown in any one of the diagrams of Figure 2 prescribes a unique relation between W, M, the maximum pressure P, and the plan size of the footing-described by some character­istic dimension L.

For any such case the dimensionless groups, which may be formed from these variables-

WL W3 W ""M' M2P ' PL2' etc.-

have fixed values. The graphs of Figures 3 and 4 show the relationship

between two of these groups,

w /w w M'\JP and PL2 '

for footings having square and octagonal plan shapes and the orientations shown.

The terms of the first group and the footing shape are the design data. Calculation of this group and reference to the graph for the footing shape specified gives the value of the second group from which the size of the footing may be calculated.

Figure 3 includes all the cases where the whole of the base of the footing is loaded as in Figure 2 (b) ; that is, where there is some pressure over the whole of the lower face of the footing. The upper limit is at the point where the minimum pressure is 95 percent of the maximum. Beyond this point the effect of the moment load may safely be neglected.

Figure 4 describes the cases where only part of the base is loaded [as in Figure 2 (d) ] . The lower limits cor­respond with cases where only about one tenth of the base area is under load. Actual designs will rarely ap­proach this condition or go beyond it. The upper limits of Figure 4 correspond, of course, with the lower limits of Figure 3.

For a given loading system on a footing, there is, except for a circular footing, some critical orientation of the axis of rotation which produces the highest maximum soil pressure. For both the square and the octagon this orien­tation is the axis passing through two vertices.

The curves for the octagonal footing have been calcu-

35

(a) (b) (c) (d) (e)

FIGURE 2-Changes in soil pressure for tall towers.

lated on this basis. Similarly, for a completely unre­strained structure on a square footing, the curves for the diagonal axis should be used for calculating the minimum size of footing.

If the structure is restrained so that rotation about only one axis is possible, as for instance in the case of a pipe rack standard, a more economical design results if a square footing is arranged so that this axis is parallel to one side. The appropriate curve is then used for calcu­lation of the footing size. In this orientation the required size of the footing is somewhat less.*

In such a situation, however, an even more economical design may result from using a rectangular footing with its greater side perpendicular to the axis of rotation.

The curves may be used equally well for design of rec­tangular footings by using the factor a, the plan aspect ratio of the footing.

a = dimension perpendicular to the axis of rotation di.tnension parallel to the axis

In 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, the graphs may be used in the same way.

The procedure then is to find the size of a square footing which, with the same loading, would produce a

p maximum bearing pressure of - This is done auto-

matically by using the groups,

marked on the graphs.

a M*a Wa - --and-- as W P PL2

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:

L ft. ft. ft.

w lb.

kips tons

M lb. ft.

ft. kips ft. ton

p

lb./ft.2

kips/ft.2

ton/ft2

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 moments in the footing itself. For convenience in these calcula-

•u : ~ :;- is less than 0. 73, the situation is reversed, but in this

cue a ~quare-or recta~footing with axis of rotation parallel to a aide should atill be wed.

36

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

minimum pressure n= --:------

maximum pressure

and in Figure 4,

p = the proportion of the width of the footing under load.

Knowing the value of the appropriate parameter and of the maximum soil bearing pressure, P, the load distri­bution over the lower face of the footing and the re­quired thickness and reinforcement may then be calcu­lated by the methods of Marshalla or Brown.1

In this connection it should be particularly noted that when a portion of the footing is unsupported by soil re­action there are shears and, more important, bending moments in the unsupported section of the slab in the opposite sense to those usually considered in the design. These stresses must be evaluated and the slab design may require modification to resist them (e.g. by the addition of top bars to resist the reverse moment) .

In common with all other methods proposed for esti­mating footing size, the calculation must be a trial and error process. The known data are usually:

• Structure deadweight (empty, working and under hy-drostatic test conditions, if required) ;

• Wind and other eccentric loads;

• Pedestal size and weight;

• Depth of footing base below ground (from knowledge of frost line level or situation of desirable load bearing strata) ;

• Allowable maximum soil bearing pressure (P).

For full details of estimation of these see Brownell and Young,2 or MarshalJ.S The moment load (M) can then be calculated from the wind load and depth of footing base below ground. The total deadweight (W), however, comprises, besides the weight of the structure and the pedestal:

• 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

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 slab thickness is of minor importance at this stage, since extra thickness of concrete only 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 separately the footing size required for several critical load conditions. These are:

• Minimum weight and maximum wind effect, e.g., in course of construction;

• Working weight a~d maximum wind;

• Test conditions-filled with water and 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 unlikely that the test period and the maximum wind would coincide. Since the construction period is normally much longer than the

T 1 T _L p ---..__ p

---f-.........._1 1 ~ ~

W = Weight of structure, footing and overburden

M = Moment of wind load and any other eccentric loads about the center line of the base of the footing

P = Maximum allowable soil be:~ ring pressure

L = Characteristic length: square-length of side octagon-width across flats

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

a = Plan aspect ratio of rectangle

_ Length perpendicular to axis of rotation Length parallel with axis of rotation

For other cases, a= 1

n (Figure 3) = Mini.mum pressure Maxtmum pressure

(See Figure 5)

p (Figure 4) = Proportion of the width of the footing which is under lo:~d.

4 5 6 8 10 20 30 40 60 80 100 200 300 400

'!!..fWD MIP

FIGURE 3-Relationship between two dimensionless numbers for all eases where some pressure acts on lower base face.

0.5

0.4

0.3

0.2

.. ,-:, ;I: 0.. 0.15

0. 10 l+++l+.<l.'tM,;' 0.09 0.08

0.07

0.06

0.05

0.04 0.5 0.6 0.8 1.5 2 3 4 5 6 7 8

~f-f FIGURE 4-Relationship between two dimensionless numbers for eases where pressure acts on only part of lower base face. (i.e. Fig. 2d).

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

The procedure to be followed is illustrated in the fol­lowing examples.

Example--Octagonal 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.2

The frost line is 4 feet below grade and the pedestal top is to be 1 foot above grade. The footing base is made

37

SIZING TOWER FOOTINGS ...

5 feet below grade, i.e., I foot belo\11 the frost line. The design ma.ximum wind velocity is 100 mph. The maximum wind moment ahout the base of the

footing is calculated to be 200,000 ft. lb. (M) .3

Tower weights are as follows: Empty Tower 30,000 lb. Appurtenances and working contents 9,000 lb. Water fill for hydrostatic test 40,000 lb.

For a pad estimated to be 13.5 feet across Oats and foot thick with an octagonal pedestal 6 feet across flats and 4 feet deep and clay fill of density 90 lb. / ft. 8, esti­mated weights are:

Concrete Fill

63,000 lb. 33,000 lb.

Then calculations for the three critical conditions arc:

Empty Working Test

w ( lb.) 126,000 135,000 166,000 M (lb. ft. ) 200,000 200,000 100,000 p (lb./ft.2 ) 2,000 2,000 2,000

:~; o.63 v63 0.675 v 67.5 t.66 v83

= 5.00 = 5.54 =15.11 w

PL2 (from graph) 0.405 0.423 0.585

L2 63 67.5 83 0.405 0.423 0.585

L (ft. ) 12.47 12.61 11.9

Thus the assumed size is too large and could be re­duced. The next trial would assume a 12-foot octagon, and the minimum size would probably be somewhere near this figure.

Example-Square Tower Footing. A square footing is to be designed to carry a total estimated deadweight of 50,000 pounds and a maximum overturning moment of 100,000 lb. ft. on soil having a maximum allowable bearing pressure of 2,000 Jb./ ft2

•

38

w~ -- - =0.5 v 25=2.5 M p

Unrestrained. Diagonal axis

w PL

2 = 0.318 (from graph)

About the Author John Buchanan is a lecturer in

chemical engineering design at New­castle University College of the Uni­versity of New South Wales, Tighe's Hill, N.S.W., Australia. He holds B.E. (Chern.) and M.E. (Chern.) de­grees from the University of Sydney. Mr. Buchanan held positions as a design engineer with Monsanto Chemicals Ltd. and Union Carbide Ltd. in Sydney prior to accepting his present position.

Buchanan

25 Lt = 0.318 = 78.6 ft.2

L = 8.86 ft.

Restrained so that rotation about only one axis is possible.

w PL

2 = 0.380 (from graphs)

25 L~=--=658 ft2 0.380 . .

L = 8.1 ft .

-0

o::, -0

Ul ·-)(

<{

a

_j

FIGURE 5-For rectangular foot ings, "a" is usually greater than one.

Example-Rectangular Footing. For the loads as in the square footing example above, assume a = 3 (short side parallel with axis):

w ~ -----M '\j P -= 0.5 , , 3 X 25 = 4.34

Wa PL~ = 0.505 (from graphs)

75 L2 =-- = 148.5 ft2

0.505

L = 12.2ft.

L - = 4.1 ft. a

The rectangle required is 12.2 ft. x 4.1 ft. having plan area 49.5 ft~ as against 65.8 ft2 for the square footing under the same conditions.

In each of the above examples the maximum pressure will be equal to the allowable, and the pressure distribu­tion may be immediately sketched after finding the value of the parameter p or n from the appropriate graph.

Considering the essentially rare and transitory occur­rence of the maximum moment load, the basic assumption stated at the beginning are sufficient for most applica­tions. The assumption of perfectly elastic soil, however, is not entirely sound and in critical cases the advice of a soil mechanics expert should be sought.

LITERATURE ClTJo:D

1 Brown, A. A. liVDROCARBON PROCI!.IISINO & P£TROLEUa.l lU.PINEil 42, No. 3, 141 (1963).

• Brownell, L. E. and Young, E. H. in "Process Equipment Design" Chapter 9, New York, John Wiley and Sons (1959).

• Marshall, V. 0. Puaou:uM lU.PtNu 37, No. 5 Dcoign Suppl. (1958).

Simplified Design Method for Intricate Concrete Column Loading

Combined biaxial bending and axial load on reinforced concrete columns present difficult design solutions. This method bypasses the usual tedious computations

E. Czerniak, The Fluor Corp., Ltd., Los Angeles

HERE's A NEW AND SlMPLlFI£D METHOD 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. It 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 visualiz­ing the problem. All given data must, therefore, be ex­pressed in 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 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.c, ..._ M,c1 A - -~-.- - -~-r-

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 Trust Computers? 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. However, when the 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 modem maiVel of technical automation '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. Hence, if he is to make the most out of the new tool, 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 purpose.

Two Design Methods. The Building Code requirements for reinforced concrete (ACI 318.56) permits columns subjected to combined bending and axial load to be in­vestigated by two methods:

• The so-called elastic method in which the straight line theory of flexure is used, except in regard to compressive reinforcement.

• The ultimate strength method on the basi~ of inelastic action.

Ultimate strength design is relatively new in American Codes, and hence some of the old timers may feel ill at ease with new concepts and new criteria. I t will be shown, through illustrative examples, that the same ap­proach applies throughout the full range, from elastic analysis to clastic-plastic and ultimate strength con­siderations. In the straight line stress distribution method. the code requires that colurflns in which the load P has an eccentricity greater than % the column depth t in

39

INTRICATE CONCRETE COLUMN LOADING .•..

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 in concrete is insignificant, and the Code requires that it be completely neglected.

Method of Analysis for Rectangular Sections. In the case of rectangular sections, it is convenient to choose one comer as the origin and let the axes coincide with two sides of the rectangle. In Figure 1: 0, B, C, and D are the comers of the given concrete section. Line QR desig­nates the neutral axis (line of zero strain), and intersects the x and y axes at a and b respectively.

Let the coordinates of the eccentrically applied load, P, be i and y; and the coordinates of any given rein­forcing bar, of area A1 be x1 and y,.

The intercept form of the equation of the neutral axis QR is :

~+..!..=t a b

Now, assuming that the stress f, at any point (x~, y,), is proportional to its distance from the neutral axis, then by multiplying fo, the stress at origin ( o, o) , by the ratio of the distances of point (x1y1 ) to that of (o, o) we ob­tain the general stress formula:

f 1 = f0 ( 1 - ~ - ~ ) The engineer need not keep track of the sign, as the

stress formula 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 tension beyond the neutral axis.

The coordinates of the centroid of the triangular area under compression, OQR, are a/3 and b/3. Hence, the average compression stress within the effective concrete section will simply be:

x=a/3

y=b/3

and the total compression load in the concrete :

ab Fe= (Average Stress) X (Area)=~ f0 X Yz ab = f0 6

It can also be shown that, for equilibrium, the load Fe is located at coordinates, a/4, b/4; hence, the moments of the compression load in the concrete, about the x and y axes are:

, _ ab a _ a2b M ••- f0 6 X T - f 024 (in x direction, about y axis)

ab2 M' 01 = f0 24 (in y direction, about x axis)

(The reader may note that a2b/24 and ab2 /24 are simply the values of the section moduli of the effective concrete area, in the x and y directions respectively.)

In Figure 1, the intercepts of the neutral axis, line QR, are shown smaller than the corresponding dimensions of the section. Hence, the effective concrete area under

40

y

I

Qc •• ,,

P<i.n

A1

(x1, y1

)

•

c

•

• 0

FIGURE 1-In rectangular sections, choose one corner as the origin.

compression, is a triangle. In the general case, when line QR is partially outside the concrete section (see Figure 3), one or more smaller triangles must be subtracted from the over-all larger one. This is illustrated in the following examples.

Reinforcing Steel Stress. The stress in the reinforcing bars is obtained by multiplying the value (fs) in the general stress formula, by the modular ratio, n, the ratio of the modulus of elasticity of steel to that of concrete. Section 60 l of the ACI Code gives the ratio, n, as equal to 30,000/f'.,.

Hence, the stress in any bar A1 designated by coordi­nates x, and y, is:

f1 1 = nfl = nf0 ( 1 - X~ - ~ )

and the load in said bar, having an area A, is:

F11 = f11 X A 1

The total load carried by the reinforcing bars (tension and compression) 1s the summation of the loads in the individual bars:

where N = total number of bars.

Similarly the steel load moments about the coordinate axes will be:

N

M'.,. = _2: F81x1

t=l

and N

M'•r = 1: F81y 1 1=1

The above formulas are completely general and the engineer, if he so wishes, may use different diameters for the individual bars, and he may or may not arrange the bars with symmetry about either axis. However, since the same modular ratio n was applied to both tension and compression bars, we did presuppose that the bond between the steel and concrete remains intact, and they deform together under stress. That is, the steel in the compression zone can withstand a stress only n times that in the concrete. In reality, this is not exactly so. Because

20"

.. - f8 ......

FIGURE 2..:...Example 1. Find eccentric load P and moments about centerline.

of plastic flow in the concrete, the compression bars are stressed more than indicated by elastic analysis. Codes 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." However, in permitting this use of 2n the Code limits the stress in the compressive reinforcing to be equal to or less than the allowable stress in tension.

In the examp les that follow, the stress in the com­pressive reinforcement shall be made equal to:

(Compressive reinforcement only)

where ft is the allowable tensile unit stress in column re­inforcement. Also, a correction shall be made for the con­crete area displaced by the steel bar by subtracting ( f 1)

from the steel stress in said bar. Hence, compressive load carried in bar A; will be:

F •• = A, (f •• -£,)= AI (2n- 1) r,

Governing Stress-Concrete or Steel Tensile. In the general stress formula for f 1, the stress at origin ( o, o) is designated as f0 , and is the maximum compressive stress in the concrete. According to the ACI Code, Section 1109 (d) : "The maximum combined compressive stress in the concrete shall not exceed 0.45fc'. For such cases the tensile steel stress shall also be investigated." Hence, f. will equal to 0.45fc' only if concrete governs, or when concrete and steel reach simultaneously maximum allow­able values (balanced design). To determine the value of f., compare:

(x/a+y!b) to I + (V0.45f'eD)

where x and y are the coordinates of the tension bar, farthest away from the neutral axis.

(a) Concrete stress governs when:

x + y f 1

7 b < 1 + .45fen hence, use: f 0 = 0.45 f' c

(b) Allowable concrete and steel stresses are reached simultaneously when:

1-r-tt...;...;....;.;.. _____ -.---;c 1211, 1s1 Az

• 117.5, ll.S)

FIGURE 3-Generally QR is partially outside the concrete section.

hence, use: f0 = 0.45 f' c

(c) Tensile stress in steel governs when:

X y ft -;-+ b > 1 + .45f'

0n

When this happens limit the stress in the extreme tension bar to ( - f 1 ) • Hence, use:

- ft f t

r. = -n-(""'"~--:---''-~--:-) = n (-;-+ ~ - 1)

Example 1. 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 in the position indicated. f' c = 3,000 psi ft = 20,000 psi.

Solution:

n = 30,000 = 10 f'c

A 1 = 0.79 sq. in/bar

hence,

r. s 0.45 X 3,000 s 1,350 psi

f 1 S 20,000 psi

a= 12" (given) 12

b - X 15=20" 12 - 3

TABLE 1--Coordinates, Stresses, Bar Loads and Moments for Example 1

&I 11 M',. )1',7 1---- I. I. I '•I I I

Poiat 11 11 12 20 (psi) ( ... ) (kipo) (la-k;po) (Ja.ki,.) - -0 0 0 ti.OO +I~ 8 0 16 0.26 +337.6 I 2.6 12.6 +0.187 +226 + 4,500 + 3.38° + 8.45 +42.~ 2 17 6 12.6 - I.Olr.l - 1462.6 -14.&26 -11.65 -202.13 -144.38 3 17.6 2.6 -D.683 - 787.6 - 7,876 - 6.22 -108.85 - 16.55 4 2.6 2.6 +O.M7 +GOO +18.000 +lUI• + 33.78 + 33.78 ------ ·---------------

Load on rtin!oreinc bars - O.S8 -268.76 - 83.t0 Load on coocrtte eft" eetive ~tetiou +63.16 +161.37 +2M.35 ---------Total +52.28 - 107.38 +172.45

• f,ow in I'Ompreuive rtln!orcemcnt ccrrte~d lor area or ooocnte dieplaeed by bar.

41

f 1 = fo (I- ~~ - ;~) Coordinates, stresses, bar loads and moments are tabu­lated in Table 1. Also see Figure 3.

The load and moments in the concrete are calculated next.

F = 1.35 X 12 X 20 c 6

0.34 X 3 X 5 3 6

k' 6

= 5 .1 lpS

M'cx= 54.0 x 3.0- 0.84x 0.75 = 161.37 in-kips

M'cy = 54.0 x 5.0-0.84 x 16.25 = 256.35 in-kips

-107.38 x= =-2.05"

52.28

. y = + 172.45 = + 3.29" 52.28

The eccentricities of the load with respect to the cen­terlines of the concrete section are:

Results:

Ex= 10.00 + 2.05 = 12.05" E1 = 7.50-3.29 = 4.21"

P = 52.28 kips Mx = 52.28 x 12.05 = 630 "k My= 52.28 X 4.21 = 220 "k

I t should be noted that in above example the ratio eft is less than 2/3 in either direction, and according to Sec­tion 1109 of the ACI Code could have been analyzed as an uncracked section. The example was selected on pur­pose, so that the interested engineer may compute, for the gross transformed section, the value of the maximum allowable load at the same eccentricities, and compare it with the 52.28 kips calculated for the assumed cracked section in Example 1.

Ultimate Strength. The term "ultimate strength design" in reinforced concrete denotes an analysis based on in­elastic action. It focuses attention on ultimate rather than design loads. As in elastic analysis, it is assumed that plane sections normal to the axis remain plane after bending, and as is common in a reinforced concrete col­umn, tensile strength in concrete is neglected. The de­parture is, that stresses and strains are not proportional at ultimate capacities. Section (A603) of ACI Code permits "the diagram of compressive concrete stress dis­tribution to be assumed a rectangle, trapezoid, parabola, or any shape which results in ultimate strength in reason­able agreement with comprehensive tests." Furthermore, it limits maximum concrete strain Eo to .003, and max­imum fiber stress in concrete to 0.85fc'. The stress in tensile and compressive reinforcement at ultimate load is limited to the yield point or 60,000 psi, whichever is smaller.

Now, when the position of the neutral axis is known or assumed, the magnitude of the ultimate load Pu and its eccentricities, which result in the prescribed limit strain, may be easily determined by using the same approach as before.

From the assumption that plane sections remain plane

42

~+..2:.._+_:_= 1 a b c

where a, b and c are the intercepts of the plane on the x, y and z axes, respectively. I t is immediately apparent that z is a measure of the strain, and the constant c is the maximum strain, which conforming to usual nota­tion, may be written as E0 • Constants a and b designate the neutral axis as before. Hence, the general relation­ship for the strain e1 at any point x., y1 may be written as:

Multiplying both sides of the strain equation by E. we obtain:

The engineer should not have any qualms about using the constant E. at ultimate strains. Since the stress in the concrete shall be limited to 0.85£0', any hypothetical stress above this value will be subtracted. The equation of the line for which f 1 reaches the value 0.85£'. may be written by making fo equal to e0 X 1000£.' (Note: Ee is assumed equal to 1000£'.). or

0.85 f' c = eo X 1000 f' 0 ( 1 - : - ~ ) (Note that f' 0 cancels out)

which for the specific case of e0 = .003 reduces to:

_x_+_Y_=l .717a .717b

from which the intercepts on the x and y axis are seen to be Xu = 0.717a and Yu = 0.717b respectively.

Stress in reinforcing bars:

( x, Yt) f 1 =nf 1---- :!Sf •

0 a b 1

About the Author Eli Czerniak is a principal design engineer with The

Fluor Corp., Los Angeles. He coordinates computer applications for the Design Engi­neering Dept., reviews manual tech­niques and develops new methods and procedures better adaptable to systems conversion in automating the design and drafting of refinery units. Mr. Czerniak received a B.S. in engineering from Columbia Uni­versity in 1949 and an M.S. in Civil Engineering from Columbia in 1950. He is a registered engineer in Cali­fornia and has published a number of technical articles. He has had field experience as a civil engineer

Czerniak and worked in design and drafting with Arthur G. McKee Co. in Union,

N. J., for two years before joining Fluor in 1953 as a structural designer. He soon headed up the structural design and drafting on various projects until assuming his present position.

z

'· '

y

' ' ' ' ' '

F .,

D

, / ,

/

FIGURE 4-This drawing helps visualize the problem in Example 2.

and, as before, a correction for the concrete area dis­placed by compressive reinforcement shall be made by subtracting the concrete stress from the steel stress, when determining the load in the compressive bar.

Example 2. Compute the Ultimate Load Pu and its eccentricities with respect to the centerlines of the sec­tion, for the neutral axis given in Example 1. Use yield point of reinforcement, f1 = 40,000 psi. Figure 4 is drawn to help visualize the problem.

Solution:

Xu = 0. 717 X 12 = 8.60" Yu = 0.717x20= 14.34"

f0 = .003 X 1000 X 3000 = 9000 psi= 9.0 ksi

Within triangle OXu Y, concrete stress equals 0.85'. = 2,550 psi

fo- 0.85 f' c = 9000 - 2,550 = 6,450 psi (maximum value of excess stress)

Load on concrete effective section

9.0x l2x20 puc= 6

2.25 X 3 X 5

6

= 221.80k

6.45 X 8.60 X 1~.34

6

and the moments about the coordinate axis

M't• = 360 X 3-5.63 X 0.75 - 132.57 X 2.15 = 790"k

M'., = 360 X 5 - 5.63 X 16.25 - 132.57 X 3.58 = 1233"k

Tabulations of the calculations are given in Table 2.

- -215 x =--- =- 1 08" 199 .

y- ~::7 =+ 4.86"

E. = 10 + 1.08 = 11.08"

E1 = 7.5-4.86 = 2.64"

TABLE 2- Loads on Steel and Concrete for Example 2

Ulti•ote Streuca I I 11 ------ M' .. M'.,

1---- II Coacrelo Steel '·I I Poi11t I I 11 12 20 ,,; poi psi kj,. 'It 'It - -0 0 0 fo·oo HOOO 2UO B 0 15 ~ +ZUO 2UO Xu 8.80 0 +2UO 2UO Yv 0 14.3. +8.283 +2650 2650 I u 12.5 .187 :!:~; 1500 + 15,000 + 10.7• + 27 + 134 2 17.$ IU -1.083 .... -40,000 - 31.8 -553 -395 3 17.$ u -Q.$83 -$~

Q ~::= -31.8 -553 - 78 • u u +0.847 +8000 +40. + 29,6• + 74 + 74 ---Loed and momenl.l oo o~l -2U -100$ -2118 Loedt and momtol.l oo coocrel.l +221.8 + 7110 +1233

Total + 188.t -215 + 887

• Loado in com.,-ve reinforcement comcted for._ of concnl.l displaeed by bar.

Hence, for the concrete section shown, a maximum ultimate load of 199 kips (divided by the proper load factor) may be placed at distances 11.08 inches and 2.64 inches from the centerline. # #

4 3

Unusual foundation

design lor • • • TALL

Close centerline distance, high towers, weak clay soil and hurricane winds gave Phillips some interesting problems

Edward V. French Phillips Petroleum Compony, Bartlesville, Oklo.

SEVERAL UNUSUAL conditions faced Phillips' en­gineers in the design of a common foundation for two tall fractionating towers. The towers were to be located in the Phillips' refinery near Sweeny, Texas.

The design conditions preSl'ntcd tht•se difficult prob­lems:

• The towers w<·n· fairly high and clost' to one an­other.

• The soil consi!'ted of a relatively weak clay.

• Horizontal forces were to be based on hurricane , winds and aerodynamic vibrations.

• Each tower had to be structurally independent of the other and each self supporting.

Layout Study. A study of proposed la>outs indicated that it was economically advantageous from a piping viewpoint, to space the· towers close to one another. An

.,..... investigation showed that for independent foundations,

TOWERS

FlGCRE 1-Gin pole ~upport~ tower "" it is raised free of ground.

44

FIGURE 2-This is the foundation after the first pour.

octagonal mats at least 40 feet in diameter would be required and that any spacing of about 40 feet or less would involve a combined foundation.

Although this investigation indicated that there would be no appreciable economy in materials using a com­bined foundation, one advantage was apparent although somewhat unmeasurable. It is possible that the towers may vibrate when subjected to steady winds of 35 to 55 mph velocity. A natural vibration period of 1.0 second per cycle was calculated for the shorter tower and 1.4 second per cycle for the taller one. Assuming aero­dynamic vibrations to occur at these frequencies, im­pulses transmitted by either tower into .a common foundation would tend to be damped by the effect of the unlike opposite tower. This damping would be effective to some degree whether one or both towers were in motion or regardless of wind direction.

In view of these factors, it was planned to space the towers on 28 foot centers using a common foundation. This spacing allowed adequate clearance for erection and maintenance operations. A plan and elevation view showing the arrangement and general details is shown in Figure 3.

Soli Conditions. The soil at the foundation site is cohe­sive. Borings were made and laboratory tests run of the soil samples. A 5-foot top stratum consists of black and tan organic clay. This is underlain with 2 feet of stiff

construction joint tie mat

and pedestal together

tan inorganic clay below which lies clayey sand and sand. Using shear strengths indicated by the tests and Ter­zaghi's bearing capacity equation, the allowable soil pressure of 4,000 psf was determined. This was based on a safety factor of 2 at a 7-foot depth with no increase permitted when combining wind and vertical loads. After the mat had been sized, uniform soil pressure due to vertical loads totaled only 1,500 psf. Computed total settlement was consequently small and a major percent­age of it could be expected to occur during construction.

Load Combinations. The effects of three separate com­binations of vertical and wind loads from the towers were investigated:

1. Vessels ready for operation plus full wind forces but without operating liquids.

2. Vessels operating plus full wind forces.

3 . Vessels ready for operation under water test con­ditions without wind.

Tower Fabrication. Schedules controlling tower fabri­cation and tray delivery were coordinated so that the trays could be shop-installed. Also, platforms, ladders and most piping were scheduled for installation imme­diately after the towers were to be erected. In addition, backfill material was to be placed before the towers were

45

erected. Since the likelihood was remote that both towers would be left stripped down for an appreciable time, no "erection" condition was considered other than to check for stability.

Wind forces were computed on the basis of 125 mph maximum gust velocity at a 30 foot height. Height fac­tors were then applied which gave the following pres­sures in three height zones:

0- 50 feet . .... . ·. . . . . . . . . . . . . . . . 52 psf 51-100 feet ..... . ... . ..... . ....... 62 psf Above 100 feet . . . . . . . . . . . . . . . . . . . 72.2 psf

These values represent pressure against flat surfaces. A shape factor of 0.6 was applied to compute the pres­sure against projected areas of cylindrical surfaces.

Anchor Bolts. In designing anchor bolts, the upper pedestals were analyzed as cantilevered flexural mem­bers, loaded with combined bending and axial forces. Compressive stress in the concrete and tensile stress in the anchor bolts was then calculated according to the theory of flexure for concrete. Both carbon and alloy steel were considered for bolt material, but investigation showed that an alloy steel with higher allowable stresses and less tendency to creep under load was the most de­sirable. The practical limit on the number of bolts that could be placed around either tower perimeter was ap­proximately 36. This in effect established the total ten­sion force which each bolt must resist. Maximum bolt diameter was not a limiting factor, but by using the higher allowable stress of alloy steel a substantially smaller bolt could be used. This was advantageous from a handling and installation viewpoint. Since some degree of aerodynamic vibration of the towers is possible, it was considered imperative that all anchor bolts be preten­sioned and that under sustained loading; elongation be held to a minimum. For each tower, 36 one-piece bolts, projecting 2 feet above the concrete, were equally spaced around the vessel perimeter. All bolts were threaded on each end and anchored mechanically .at the bottom with a 2Y2-inch thick rectangular plate held between two torqued nuts.

Octagon Pedestal. Because vertical loads for both towers were considered equal they were centered sym­metrically about the foundation centerline. The condi­tion causing the greatest eccentric loading from vertical forces alone would result from either tower being water tested singularly. This eccentricity, found to be consider­ably less than that caused by maximum wind forces, was not critical. The combined wind overturning moment from both towers applied at the top of foundation was 27,750 foot kips. Any wind shielding effect by either tower was neglected and the overturning moment was assumed equal in all directions. Because of this, an octagonal outline for the foundation mat was more suit­able for limited soil pressure than was a square or rectangular shape.

Using a maximum toe pressure of 4,000 psf, a 50-foot diameter octagon was found to satisfy all load combina­tions, with the number 2 load combination actually con­trolling the diameter. The weight of the operating liquids was relaively small when compared with the total mass and overturning moment. As a result, there was less than 10 percent difference in the toe pressure and/or

46

11 •12,1701"

10' 'to. X 203'

J)_

PLAN

14' I· 14'

ELEVA nON

" I

11•15,5801"

FIGURE 3-Plan and elevation showing towers spaced at 28 feet on centers.

DESIGN CONDITIONS

One tower is 11 ~ feet in diameter by 177 feet in height. The other is 10 feet in diameter a.nd 203 feet high. Although the towers differed considerably in size, there was less than 3 percent difference in the cal­culated vertical loads for each. This was found to be true for both operating and empty conditions. For design purposes, vertical loads for each tower were considered equal. Empty tower weights included the v~ls plus all accessories ready for operation. Oper­ating weights consisted of empty tower weights plus operating liquid. Other data and conditions which governed foundation design are as follows:

Tower weight, empty, each 425 kips.

Tower weight, operating, each 500 kips.

Maximum velocity of wind, 125 mph.

Maximum allowable soil pressure, 4,000 psf.

Maximum settlement allowed, ~ inch.

Minimum stability ratio, 1.5.

Concrete-3,000 psi in 28 days. Where applicable ACI Code {318-56) to govern desiP, and detailing. f0 and f1 to be increased by one-th~rd where stresses are due to combined wind and vertical load.

Maximum allowable anchor bolt stress:

Alloy steel, 40,000 psi.

Carbon steel, 20,000 psi.

Unusual Foundation Design .. .

eccentricity between operating and empty conditions. The stability factor under load combination 1 was 2.4, and for load combination 2, 2.6.

Step Sedion. After the mat had been sized it was de­termined, by trial and error calculations, tha~ a stepped section through the ~enter was desirable. Thts step was run continuously one foot thick, across the mat center and for convenience w.as made equal in width to the octagon side. At the edge of the step, the mat depth was set at three and a half feet which with sufficient bottom steel would approach a balanced design for the resist­ing moment from soil pressure. This d.epth was th.en continued to the outer edges of the mat m order to mm­imize the steel requirements and to maintain over-all stiffness. The pedestals were then made equal in width for symmetry and connected: A minimum allowable cover of 12 inches outside of anchor bolts on the 11-foot 6-inch tower 'determined the 14-foot, 6-inch width.

' ' Mat Reinforcing. The mat reinforcing in the transverse direction or perpendicular to face of the step w~ deter­mined by analyzing sections across the ent!re Width of the foundation. Shear and moment at sections through the center, at the face of pedestal, at the face of step, and at points between the step and outside edge of m~t were computed. Load combination 2 caused a maxi­mum moment at the face of the pedestal and step and at other points toward the edge of mat. Maxim~m mo­ment in the same direction through the foundatiOn cen­terline was caused by load combination 3. To satisfy this shear and moment, eighty-one # 11 bars were spaced on six inch centers in the bottom of the mat at the face of the pedestal and step, forming a center strip 40-feet wide. Alternate bars of the above group, plus three shorter #11 bars along each edge were extended through the center to the opposite side totaling 4 7 # 11 bars to resist moment through the cener. As the moment decreased toward the outer edge of mat, alternate bars were discontinued in two stages leaving # 11 bars on 2-foot centers at the extreme outer edges.

In computing transverse reinforcing req~irements ~or the top of mat, negative moment on the wmdward s1de caused maximum tension at the face of the step. Here, 41 # 11 bars were placed on one-foot centers with al-

About the Author

Edward V. French is a senior structural design engineer with Phillips Petroleum Company, Bar­tlesville. lie directs the structual and civil engineering design phase of assigned projects. Holder of a B.S. deg ree in ~ivil engi~eerin~ from the Univers1ty of M1ssour1. He has been with Phillips in the Engineering Department since his graduation in 1952. Previous to this time he had two years' expe­rience in general construction work.

ternate bars discontinued in two stages both toward the outside and toward the center, leaving # 11 bars at 4-foot to run continuously through the center. In com­puting this moment, only the weight of the .overburden directly above the mat, plus the concrete m the mat, was considered acting downward.

The heaviest reinforcement in the opposite direction or parallel to the longitudinal axis of the pedestal was also required for load combination 2. Assuming the wind from a direction parallel to the longitudinal axis of the pedestal, tension from the wind moment on the leeward vessel combining with the effect of soil pressure pro­duced maximum tension in the top of pedestal. This was near the inside face of the leeward tower. Assum­ing all the tension to be resisted by longitudinal ste:l alone 22 # 11 longitudinal bars were placed for th1s purp~se in the top of pedestal. For the same combina­tion of forces tension in the bottom of the mat near the inside face ~f the windward tower required 47 # 10 longitudinal bars. These were placed in a 20-foot wide strip through the center of foundation. Other reinfo~ce­ment in the longitudinal direction was of a nommal nature and was placed in sufficient quantities to assure proper continuity.

Shear Key. It was first considered desirable to specify a continuous concrete pour between the mat and pedes­tals thus providing the best possible shear connection between the sections. Several factors making a continu­ous pour impractical were excessive .dead l~~ds ~n forms; inaccessibility; possible difficulty m pos1t10mng bolts; and unnecessary exposure of the excavation to weather. A large portion of the anchor bolts and pedestal rein­forcing totaling some eleven tons would normally re­quire support from pedestal form work and create a support problem. Concrete placement in the center portion of the mat would be difficult with all pedestal reinforcing and bolts in place. It was felt that accuracy in positioning anchor bolts might be sacrificed if a con­tinuous pour was made. Assuming a continuous pour, the excavation would be exposed to weather longer be­fore pouring could begin, the~eby subjecting .the s?il below the footing level to detnmental change m mms­ture content.

In order to eliminate these disadvantages, a construc­tion joint was designed between the step and pedestal so that the mat and step could be poured first. A 14-foot, 6-inch wide by l-inch deep recess centered beneath each tower provided a four way shear key between step and pedestal. This recess also provided additional depth for maximum bolt anchorage.

Vertical Reinforcing. Particular attention was given to the selection of adequate vertical reinforcing through the center of foundation, tying the mat and pedestal .to­gether, because of the unusually high vessel overturnmg moment. The two, fourteen and one-half foot octagons were first assumed to act as separate round stems and the connecting center section neglected. They were then analyzed as round sections acting in bending and dir~ct stress the critical section being taken at the construction joint.' This analysis resulted in a total of 120 square inches of vertical bars required for each stem. Under this assumption, these stems could tra~sfer ~l of the over-turning moment from the towers wtthout mfluence

47

from the connecting center section. This connecting sec­tion became functional when the full depth of the foun­dation was considered a flexural member resisting a mo­ment in the transverse and longitudinal directions.

For vertical bars in each stem, 120 #9 bars were arranged into two rows, one row on either side of the anchor bolt circle. It was felt that in placing these bars in two rows, stress from the anchor bolts would be trans­ferred more evenly and that any tendency for the con­crete to separate at the construction joint would be min­imized. Additional #9 bars were then spaced on 1-foot centers along each side of the connecting center section to prevent separation at the joint when the entire foun­dation acted in flexure. Figure 2 shows the foundation after the first pour was completed and it also shows the vertical bars and the keyed construction joint used to tie the mat ;tnd pedestal together.

Gin Pole Bases. The possibility of combining a con­crete base, which would support and anchor the tower erection gin poles was considered. By providing such supports, considerable time and labor could be saved when setting the poles by eliminating the need for tying down the pole bases. The position of each tower prior to raising was planned with the tower lying at 45" to the main foundation axis. The pole bases at the closest possible position would straddle either tower on approx­imately 30-foot centers. Figure 1 shows the poles with the 10-foot by 203-foot tower free of the ground sup­port. When the foundation was analyzed, applying con­centrated vertical reactions from the poles spaced at 30-foot centers, it was found that tension across the top of the concrete might cause extensive cracking. This cracking, although probably not detrimental, was un­desirable and to prevent it, additional heavy reinforce­ment would be required. The estimated additional cost of materials to provide these integral foundations was estimated at $2,000. This was considered too costly for the advantages offered and the plan was abandoned. As an alternate method, the gin pole bases were set outward and placed on timber cribbing completely clear of the tower foundation with cables providing the necessary anchorage.

Leveling The Towers. The pedestals were poured to within 2 inches of the finished elevation. As the towers were erected, the base rings were set on a series of steel shims which had been previously leveled. Final leveling of the towers was then accomplished by adjusting shims and anchor bolt nuts. After all adjustments were com­pleted, two inches of grout was placed across the top of pedestal and beneath the tower base rings. Each an­chor bolt was torqued to an initial stress of 45,000 psi. No inconvenience was reported by the contractor be­cause the anchor bolts projected two feet above the con­crete. Neither was there any difficulty reported in re­gard to spacing anchor bolts to match the tower base rings.

Materials. A total of 383 cubic yards of concrete and 27 tons of reinforcing steel was placed in the foundation. As mentioned previously, no large savings, if any, in concrete materials were realized over those required for

48

FIGURE 4-Tower installation complete with insulation, platlorm and piping.

two independent foundations. This was substantiated by further experience when a third tower of similar propor­tions ( 10~ feet x 177 feet) was designed and installed simultaneously, nearby. This tower was placed on the usual mat and pedestal octagonal foundation and re­quired only 140 cubic yards of concrete. However, under the circumstances which established the design condi­tions, there were still advantages in the saving of space in conformance with the best piping arrangement and in the possible vibration damping effect gained.

Trend. There is a definite trend in the industry toward the use of taller fractionating vessels containing more trays. The experience acquired during the design and installation of these towers will be useful in determining the feasibility and planning of future units. ##

NOTES

49

FIGURE 1 FIGURE 2 FIGURE 3

Foundation Sizing Simplified Tables can be used to select foundations as easily as capacity tables are used to select pumps

David H. Kannapell, Girdler Construction Corp., Louisville, Ky.

JUST AS a designer can select a storage tank using capacity tables, so a structural designer can choose a foundation based on tables of capacities. Entering the tables with a given weight and moment (or eccentricity) you can quickly select the minimum size of foundation required. In addition, you can readily determine the distribution and magnitude of soil pressures under the foundation.

How to Make Capacity Tables. Two cases of foun­dation loading are considered in developing capacity tables. The first case consists of a loading which pro­duces uplift on part of the foundation. This case is shown in Figure 1. The second case covers bearing under the entire foundation and is shown in Figure 2.

For calculating the capacities of an octagon founda­tion subjected to soil pressures as shown in Figure 1, the following formulas1 are used.

C,=7r/8(1 - 2k) + ( + - +k++k' )~~+

50

. ( 2k- 1) arc sin (2k- 1) 4

( 5 7 k 1 k' + 1 k' ) C,=7r/64 (5-Sk) + 16 - 24 -6- 3

/ k k' + (Sk- 5) arcsin (?k--1) \j - 32 -

2(1-k)

E ·=D X (2C,-C,) • 2C,

M = EP

The capacities of an octagon foundation subjected to loading as shown in Figure 2 are determined from the following formulas'.

E=~X (1 - m) 8 ( 1 + m)

P=8E2p X (l+m)' ( 1-m)'

M= EP

NOTATION

P = Concentric vertical load capacity, kips

M = Overturning moment capacity, foot-kips

E =Eccentricity of load to produce corresponding moment, feet. E = M/P.

p =Unit soil pressure, kips per sq. ft.

k = Ratio of unloaded length of diameter to diameter of in­scribed circle of octagon. Used in Figure 1 loading only.

m =Ratio of minimum unit soil bearing to maximum soil bearing. Used in Figure 2 loading only .

D . = Diameter of a circle equivalent to the inscribed diam­eter, D, of an octagon, feet. D.= 1.04D.

C, and C, are coefficients used to shorten algebriac opera­tions.

By decrementing k and incrementing m, capacities of a one-foot diameter octagon are developed based on a maximum unit soil pressure of one kip per square foot. The relations of the capacities and eccentricities of any other diameter octagon, n., to those for the one-foot octagon are as follows:

Ex ·= E,._,;,, X Dx for a given "k" or "m" value

P. = P, • ....,, X n.• for a given "k" or "m" value

M. = M, • ...,, X n.• for a given "k" or "m" value

Tables 1 and 2 are illustrative of tables that may be used to estimate and design footings subjected to the forces described. The tables were developed on an elec­tronic digital computer.

ILLUSTRATIVE PROBLEMS

To demonstrate the use of the capacity tables, several illustrative problems arc presented as follows:

Problem 1. Size an octagonal foundation for the self­supporting vertical vessel shown in Figure 3, using the following data:

p = 2000 lbs./sq. ft. (2.0 kips/sq.ft. at 4'-0'' below grade)

P. =50 kips, weight of tower

M = 147 ft.-kips, about base of tower.

Since the tables show capacity of foundations based on a maximum soil pressure of 1 kip per square foot, it is necessary to first use an "adjusted" value for P and M to compensate for the larger soil bearing value. The fol­lowing relationships are used :

I

k

m

"Adjusted" P = ""'7':-:----.::.P_-:-:-~:---::-­Allow. pressure, kips/sq.ft.

TABLE 1-octagon Diamete r = 11.0D

korm p M E

ru 37.009 85.596 2.312 .20 40.037 84.784 2.117 .15 43.016 !'3.016 1.929 .10 45.923 80.412 1.751 .05 48.729 77.147 1.683 .00 61.393 73.493 1.429 .05 53 .. 963 69.818 1.293 .10 66.533 66.143 1.169 .15 59.102 62.469 1.066 .20 61.672 58.794 .953 .26 64.242 55.119 .857 .30 66.812 61.445 .769 .35 69.381 47.770 .688 .40 71.951 44.095 .612 .45 74.521 40.421 .642 .50 77.090 36.746 .476

r 55 79.680 33.071 .415

60 82.230 29.397 .357 65 84.799 26.722 .303 .70 87.369 22.047 .252 .76 89.939 18.373 .204 .80 92.609 14.698 .158 .85 95.078 11.023 .115 .90 97.648 7.349 .075 .95 100.218 3.674 .036

Capultlee hued on 1000 lb/ aq. f t . ano ... b le ~u bearing

Area of baae • 100.2 aq. ft.

T h lckneu Weight (klpe)

1'---Q* 15.0 1'--6* 22.5 2'-o• 30.1 2'-6"' 37.6

"Adjusted" M = ---;-;-:-----'M:..::...__,.-:---:-~­Allow. pressure, kips/sq.ft.

The eccentricity, E, remains the same for any soil pressure.

Step 1- Assume weight of foundation, pier and earth backfill as 50 kips.

P = 50.0 + 50.0 = 100 kips.

Step 2-Calculate eccentricity and "adjusted" value of P:

"Adjusted"

E = 147 = 1.47 ft. 100

P _I00.0 _ 500k' ---- . tps 2.0

Step 3- Enter tables with "e" and "adjusted" P. Select 11'-0'' octagon (Table 1) (table capacity P = 51.393 kips, "E" = 1.429 ft., k = 0.0 ) .

Step 4-Check assumed foundation weight:

Pier: 5'-0" Oct. 3'-6" high = 10.87 kips Foundation: 11'-0" Oct. 1'-6" thick = 22.50 kips Backfill: (100.2-20.7)x 2.5x0.1 = 19.88 kips

Total = 53.25 kips

P = 50.0 + 53.25 = 103.25 kips

E = 147.0 = 1.42 103.25

Adjusted P = 10~·25 = 51.6 kips

Re-enter Table 1 and check selection. Inspection indi­cates 11'-0'' octagon is satisfactory. Since k = 0.0, the distribution of soil pressure is such that 100 percent of the foundation is under compression; minimum soil pressure is zero on the windward edge and 2,000 pounds per square foot on the leeward edge.

Alternate Solution.

Step 1-Same as in original solution.

TAILI 2-octagon Diameter = 14.0D

korm p M E

r5

59.949 176.466 2.943 .20 64.8M 174.793 2.695

k .15 69.679 17U47 2.466 .10 74.387 165.778 2.228 .05 78.933 159.048 2.014 .00 83.249 151.514 1.819 .05 87.412 143.938 1.646 .10 91.574 136.362 1.489 .15 95.737 128.786 1.345

1 ·20 99.899 121.211 1.213 .25 104.062 113.635 1.091 .30 108.224 106.059 .979 .M 112.386 98.484 .876 .40 116.549 90.908 .779 .45 120.712 83.332 .690

m .50 124.874 75.766 .606 .65 129.0.16 68.181 .528 .60 133.199 60.605 .464 .65 137.361 63.029 .386 .70 141.524 45.453 .321 .75 146.686 37.878 .259 I .SO 149.849 30.302 .202 .85 154.011 22.726 .147 .90 158.17-l 15.150 .096 .95 162.336 7.575 .046

Capacltlee Baaed on 1000 lb./ aq. ft. Allowable Soli Bearing

Area of Baae - 162.3 aq. ft.

Thlckneea W•laht (klpa)

1'-o" 24.3 1'-6· 36.5 2'-o• 48.7 2'~· 60.9

51

Foundation Sizing Simplified . ..

Step 2-0btain "adjusted" values of P and M:

"Adjusted" P = ~~~0° = 50.0 kips

"Adjusted" M = 1::0° = 73.5 ft.-kips

Step 3-Enter Table 1 with these adjusted values. Se­lect 11'-0" octagon as bef~re.

Step 4-Same as in original solution.

Problem 2 . This problem illustrates the method for obtaining sizes of foundations other than those given in the tables. For this problem refer to Figure 3 and use the following data:

p = 5,000 lb.)'sq. ft. (5.0 kips/sq.ft.) at 4'-0" P.= 235 kips, weight of tower M = 990 ft.-kips

In the solution of this problem, the following rela­tionship is used:

D,' =_f.!. or D, = - /_f.!. X D' D' P V P

for a given "k" or "m" value.

Where D1 = octagon diameter desired D = table octagon diameter P1 = load to be carried by octagon D1

P = table octagon load capacity.

Step 1-Assume weight of fou ndation, pier, and earth backfill as 85 kips.

P = 235.0 + 85.0 = 320 .kips.

Step 2-Calculate eccentr icity and "adjusted" value of P :

E - 990.0 = 3.09 ft -320.0 .

"Adjusted" P = 320·0 = 64.0 kips 5.0

Step 3-Enter Table 2. Closest capacities are P = 59.949 kips, E = 2.943 ft., k = 0.25 for 14' -0" octagon. P = 68.819 kips, E = 3.153 ft., k = 0.25 for 15'-0'' oc­tagon.

D = y 64.0/59.949 X 14.0' = 14.47 ft., say 14.5 ft., and

E = 14.5/14 X 2.943 = 3.06 ft.

Step 4-Check assumed foundation weights: Pier: 5'-0" octagon 3'-6" high Foundation: 14'-6" octagon 1'-6" thick Backfill: (174.1-20.7) x2.5xO.J

Total P= 235.0 + 88.47 = 323.47 k,ips

E=~=306ft 323.47 . .

"Adjusted" P = ~ = 64.69 kips 5.0

D = ~ 64.69/59.949 X 14.0' = 14.62 ft.

= 10.87 kips = 39.20 kips = 38.40 kips = 88.47 kips

Since k = 0.25, it is immediately known that 25 per­cent of the diameter of the octagon is unloaded and 75 percent is loaded; the unit soil pressure varies from 0 on the windward side to 5,000 lbs. per sq. ft. on the lee­ward side over the loaded length.

52

Problem 3. This problem illustrates use of tables to determine soil loading under an existing foundation. For this problem, refer to Figure 3 and use the follow­ing data:

P. = 300 kips, weight of tower M = 373 ft.-kips D = 11'-0", diameter of octagon

In the solution of this problem, the following relation­ship is used :

~=~or p, =~for a given "k" or "m" value. 1.0 p p

When p1 = actual maximum unit soil pressure, P1 = total vertical load, P = Table 1 octagon load capacity based on 1.0

kips per square foot soil pressure.

Step ! -Calculate weight of pier, foundation, and earth back-fill:

Pier: 5'-0" octagon 3'-6" high = 10.87 kips Foundation: 11'-0" octagon 1'-6" thick = 22.50 kips Back-fill : ( 100.2- 20. 7) x 2.5 x 0.1 = 19.88 kips

Total = 53.25 kips

Step 2-Calculate total load and eccentricity:

Total load = 300.0 + 53.25 = 353.25 kips.

E = 373·0 = 1.06 ft. 353.25

Step 3-Enter Table 1 for 11'-0'' octagon with known "E." Read P = 59.102 kips.

m=0.15.

p. = 353.25 = 5.98 kips/sq. ft. (5,980 lbs. per sq. ft.), 59.102 on leeward edge.

M inimum soil pressure mp1 = 0.15 X 5980 = 897.0 lbs. per sq. ft. on windward edge. 100 percent of foot­ing is under compression.

The method outlined herein has been limited to oc­tagon foundations for brevity. Using appropriate for­mulas2, the same method may be applied to square foundations with an overturning moment abou t both the rectangular and diagonal axes.

LITERATURE CITED 1 Fork, Chas. A., "Graphical Methods Aid io Stack Foundation De­

sign" Petroleum Rdiner SO, No. 3, p 81 (1951). 'Fork, Ch .... A., "Applying Graphical Methods to Square Footing

Design" Petroleum R~6ner 31, No. 11, p 145 (1952).

About the Author

David H. Kannapell is a senior structural engineer with Girdler Construction Corp., Louisville, Ky. where he performs civil and struc­tural design of gas processing and chemical p lants. Holder of a B.S. degree in civil engineering from the University of Louisville ( 1936), Kannapell has had structural de­sign experience in many large chemical plants, synthetic am­monia, hydrogen production, gas purification and carbide manufac- D. H. KannapeiJ turing plants. He is currently the president of the Louisville chapter of the Kentucky Society of Profes­sional Engineers.

Dowel Sizing For Tower Foundations

Tower pedestal dowel bar reinforcing is usually oversized by a commonly used formula with a high safety factor. A more economical method is presented

Andrew A. Brown, Union Carbide Chemicals Co., South Charleston, W. Va.

ffilll!,~ EEE: ~111111 -+-'1

K

1.00 .95 .90 .85 .80 .75 .70 .65 .60 .55 50

.45

.40

.35

.30

.25

.20 .1 5 .10

1 ~

f* ' i I

I= I== 1 lf. •o = I= : >-;-

I= I= It · , 11=

t;; t -~ r"·

--:::. :)::::.!: }!;

1:±: Itt 1-'- = ~ 1'-' ··:o ff

[i I~ ~

~

" ;: t T

f ' ,...._

THE SIZING OF THE DOWEL reinforcement is usually the last part of tower foundation design. All combina­tions of loads and moments are requiied in computing the base slab. The pedestal size is usually fixed by the base ring and anchor bolt spacing. So, with these data, the size and spacing of the bars can be assumed for analysis. Some designers use a minimum percentage of the area of concrete for the reinforcement similar to concrete column design practice.

Example. As an example, the following design data of an existing column will be used:

M = 7,000,000 inch-pounds, the maximum moment

lJE 1m tffil lli= b

= ~ "l

.. 1!.

'H I ~

·~ :~ 1£

;; r :r. £ ' :t IE ~~ :

:;: :

· k ..

-

I! h lffipll{~ 1cv _;_ u.J ::.~ = t:l1tt

! fl

-i!J I ~

1: 11 ::1 If-I

l_ :.:_ i-'- 'j_ :j ; 1 l1 ' $ .j: f' ~ ill ~h,;

: ·1z i': r '] l:!: I '~ IJ ci_ Ak1 h . Jt lt.:ii'=H + v. .t : 11! lf ~ :Jll"f_ ~::J;E

.i .. !ffilifrl!f iEl[ffi E: '.4i ru: IRE jffil Jl!, j,; g

1,1; FEE IW:'I l iiE lim l[ : .rn; [

-if lf_ Ill f-iT r'f;i 1:! IE i+ t4j It 1:1 !i lil I til ~ ; jj_J lt: t-t

o<:

180° 154°09' 143°08' 134"26' 126°52' 120° 113"35' 107"28' 101° 32' 95"44'

90' 84°16' 78"28' 72°32' 66°25' so• 53"08' 45"34' 36"52'

CM- .10 20 .30 .40 .60 .80 1.0

.06 .08 .10

2.0 3.0 4.0 6D 8.0 0.0

.60 80 1.0

20.0 30.0 40.0 - C M

cv - .03 .04 .20 .30 .40 2.0 3.0 4.0 6.0 8.0 10. -cv FIGURE !-Coefficient curves used to find unit stress.

53

DOWEL SIZING FOR TOWER FOUNDATIONS ...

P = 80,000 pounds, the minimum load which includes the weight of the concrete pedestal

r = 36 inches, the radius of the inscribed circle R = 33 inches, the radius of the dowel bar circle Reinforcement: 20 number 8 bars, NA8 = 20 (.79)

= 15.8 square inches n = 10

With this information t.he~is computed, k values are r

assumed and the various determinations made until the

neutral axis is located. Then, the ~f the internal r

stresses equals that of the external forces. As a convenience in recording the values, a table is

constructed .. This can be revised to suit the individual. The analysis follows:

1) e M 7,000,000 = 2.43

80,000 (36)

SM = 24pn7T - = - = 2.45 2 ) (R)2 24(10)(20).797T ( 33 )2 r 7T(36} 2 36

3)

3 n7T = 3 (10) 7T (20) (.79) = .

366 p 7T (36) 2

Now try k = .26, from Figure 1, CM = 5.1 and CV = .41 (see table below for complete investigation which shows efr = 2.01 or too small.) The other k values are tried until the e/ r approaches 2.43 ("k's" of .25 and . 245 brackets this e/r).

CM+SM Note: SV = 37Tpn ( l-2k) and-;- = 16(CV _ SV)

CM+ k CM SM SM (I-lk) 3wpn sv cv 16(CV-SV)

- ------ --.26 6.10 2.45 7.65 .48 .366 .175 .410 3.76 .23 3.90 2.45 6.35 .54 .366 .198 .310 1.79 .24 4.30 2.45 6.75 .52 .366 .190 .340 2.40 .2/i 4.77 2.4.~ 7.22 .50 .366 .183 .380 3.15 .245 4.1\7 2.46 7.02 .51 .366 .187 .360 2.77

e/ r

2.01 <2.4 3.5.5> 2.4

3 3 3 3 3

2.81> 2.4 2.29 <2.4 2.54> 2.4

The unit stress m the concrete can now be computed . 96kM

for these two k's by usmg formulas fc = (C + for M SM)r3

nf0 [R+ r(1-2k)] concrete, and r. =

2kr

for the reinforcement. (Equations 5 and 7)

f - 96 (.25) 7,000,000 - . -c- 7.22 (36)8 - 497pst f.-

(497) 10[33 + 18]

18 = 14,100 psi

f = 96 (.245) 7,000,000 c 7.02 (36)3

502 psi r. =

(502) 10 [33 + (36) .51] . 2 ( .245) 36 = 14,600 pst

By comparison with the conventional method of 4M P NO - N , we get the force imposed on the maximum

4 (7,000,000) 80,000 stressed bar =

20(66

) - ---w- = 21,200-4,000 =

17,200 pounds.

. . 17,200 The umt stress IS Q.79" = 21,700 psi.

54

It is apparent that the latter solution is not very economical, and contains a factor of safety out of propor­tion to the other elements of the foundation. Its use should be discontinued.

Derivation of Equations The subject of foundation design for tall stills and towers

has been accorded much thought during the past year as evi­denced by numerous articles. Other equally important items such as anchor bolts and dowels have been of less concern. Most writers subscribed to the use of the approximate, inaccurate and

uneconomical formula of ~~- ~ for this purpose. As expressed

by many, it is safe. Actually it provides a factor of safety out of proportion to the other designed elements and is merely an expedient. One would not dare to oversize the other parts of the structure proportionally as he would never be retained for a repeat performance.

To stimulate and provoke thinking toward the development of a more rational analysis for dowel bars, this method ts sub­mitted. It is not presented as the final answer but with the hope that it will influence others to produce something better for our use.

For this presentation a cylindrical pedestal, or that formed by the inscribed circle of the octagon or other regular polygon is used. The working stress design method is employed with the attendant assumptions. A section that is plane before bending remains plane after flexure is imposed. Stress and strain vary as a straight line and directly as the distance from the neutral axis. The reinforcement takes all tensile stress due to flexure.

In the development of the formulas the reinforcement is re­placed with an area of Es/ Ec times that of the steel. In con­structing the transformed section, the holes in the concrete were not removed from the compression area. This should have very little influence on the end results and is partially neutralized by the area outside of the inscribed circle. It does simplify the derivations considerably .

The symbols used arc the same as those usually found in concrete design manuals and text books employed for teaching this subject.

Figure 2 shows a typical foundation with the forces acting on it and gives the location of the dowels. A section is taken through the pedestal just above the foundation slab and the forces acting on this section are located in Figure 3.

The equations representing the total forces and moments im­posed on the concrete and reinforcement are now derived.

Taking the summation of moments about axis Y-Y we have M- M

0- M 8 = 0 or M = M 0 + M 8 • By summation of the

forces in the Z direction we get P-( V c + V 8 =0 P = V 0 +V •.

The total vertical force acting on the concrete is the sum of all the stress acting on the segment of the circle to the right of the neutral axis.

If fc' represents the intensity of stress on the elemental area dA

0, then dV0 = f0 'dA0 , dA0 = 2ydx = 2rsin</> ( rsin</>d</>} fc' =

fc (cos</>-cosa) B b . . h I . h bo h y su stttutmg t ese va ues m t e a ve; t e (1-cosa)

2f r2 )a total force acting on the concrete, V 0 = c (cos 4> -

(1- cosa) 0

cos a) sin2 4> d </>.

Integrating and substituting a for 4> and 2k for ( 1-cosa) we

f r2 [sinS a sin a cos2 a- a cos a] obtain V e = T -

3- +

2 , Equation I

Taking moments of the internal stress in the concrete about axis Y - Y, it follows that dM

0 = xdV c· The moment of the

force f.' on elemental area dA0 about Y - Y becomes dM0 = f.' dA

0 rcos 4> as x = rcos </>. Substituting the values of dA0 and

fc' as before, the total moment becomes

2{ r'l jCI M

0 = e (cos 4>- cos a) sin2 4> cos 4> d 4>

(1 - cos a) 0

2£ rs [ = (

1 c ) - Y8 ( Y. sin 44>)

-cos a

cos a sins 4> ]« 3 0

= {0 rs [a + cos a sin a- 2 coss a k 8

_ cos a sinS a] 3 '

Equation 2.

The total force V 8 acting on the steel is found by converting the dowel reinforcement into an annular ring of equivalent area of concrete and of width t. The width is equal to the product of n and total area of the dowels divided by 2 'iT R . Let f" equal to the intensity of stress acting on an area dA

8 which is located

a distance of R cos<{> from axis Y-Y. Then dV1 = f"dA8 = f" t Rd </> By similar triangles

f" R cos</>-r cos a f0 (R cos</>- r cos a) {" = r; - r(l- cosa) r ( l-cosa)

. . 2ft R By substitutiOn dV8 = ,, .,..,.--"-0--7

r(l - cosa) (R cos</> - r cos a) d <{>

( .... then V = 2f• tR

• r ( l- cosa) )o (R cos</> r cos a ) d <f>

= 0 t R sin <{>- r ~ cos a 2f R [ ] " "" r(l-cos a)

0

2f0 tR -:-:----"-----.,.. (- 'iT cos a), since A

8 = pn 'iT r2 = 2 'iT Rt

r (l-cos a)

2Rt = pnr2 and (1- cos a)= 2k then

{0pnr2 'iT cos a . v. = -

2k , Equation 3.

The moment of the forces acting on the dowel bars about axis Y-Y can be obtained by getting the summation of the mo­ments of the forces acting on all the small dA areas. dM

1 =

dV8 (Rcos</>)

2 f tR2 c (R cos<{>-rcosa) cos<{>d</>

r(l-cosa)

2 f tR~ } '"00

M.= • (R cos</> - rcosa)cos<{>d<f> r(l - cosa)

0

= 2

£. tR2

[ R <t> + R sin</> cos<{> -· r cos a sin</> ]0

1100

r(l-cosa) 2

Substituting the limits, Rt = p~~ , and 2k = ( 1-cos a) we get

f pnrR2 'iT M = 0 Equa tion 4. 8 4k

About the Author

Andrew A. Brown is a structural engineer with the Union Carbide Chemicals Co., South Charleston, W. Va. His work at Carbide in­cludes the preparation of structural designs, reports and analyses for all types of frames and foundations both new and existing. Mr. Brown's profes~ional experience includes that of a bridge consultant with 12 years active duty in the U.S. Navy Civil Engineering Corps as a public works officer and 10 years in the Bridge

Brown Dept., State Road Commission of West Virginia. He holds a B.S. de­

gree in civil engineering from the University of West Virginia. H e is a member of the Society of American Military Engineers and Tau Beta Pi.

I FOUNDATION

M. = e p

FIGURE 2- Typical tower foundation showing dowel loca­tions.

X

c -:n=

II

y

X

dAc= 2r2sin2¢ d¢ {2 ydx)

r!cos¢- COS.c) --f.-+o,__-++---t--

Mc+ Ms --- =e Vc t V5

r--.+f'-+....;R~c""'o,;:s ¢- r c o s ...

Z 2 kr = r{ 1- cos.c}

SECTION X-X AXIS

X

~~

FIGURE 3-Section A-A through Figure 1 pedestal just above foundation slab.

55

DOWEL SIZING FOR TOWER FOUNDATIONS ..

Now Equation 1 and 3 are added and multiplied by "r''

r (V., + V 0 ) =

f.,rS (sin' a + sin a cos2 a - a cos a k 3 2

f.pnrS., cos a) 2k

= f•r3 (2 sin* a+ 3 sin a c•os2 a- 3 a cos a - 3 pn 'TT cos a) 6k

This is the product of r and the total streucs in the concrete and reinforcement and equals the external load P x r. The total moment of internal stresses is M0 + M

8 =Equation 2 +4

TAILE 1--c:alcvlated values of numerl~al ~oefflclents CV and CM

K

.10 ....•....•..•....•..•..•.

.15 ...... " .............. ..

.20 ... " ........... " ... ..

.25 ..............• . ...•.....•.

.30. . ......... . ........... .

.35. .. .. " .............. ..

.40. .. ........... ..

.45. .. ...• " ............. ..

.50. . ...... . .............. .

.55.. .. ................. .

.60. • ............. .

.65.. • ................... .. .70. .. ................. . .75 ................... . .80. • •.•••• • •• ••.• .85. .. ................ .. .90. • ................... .. .95. . .............. ..

1.00 .... ". . • • ...... .

" 36° 25' 45° 34' 63° 08' 6()0 oo• 25' 7ZO 32' 78" 25' 84° 16' 000 95° 44'

101° 32' 107" 28' 11:!0 35' 120" 126° 52' 134° 26' 143" 08' 164° 09' 180"

cv .0. .11 .22 .38 .59 .85

1.18 1.56 2.00 2.50 3.06 3.68 4.36 5.10 5.87 6.71 7.58 8.49 9.42

CM

.57 1.51 2.91 4.77 7.05 9.67

12.58 15.77 18.85 22.00 25.12 28.03 30.65 32.93 34.79 36.19 36.92 37.51 37.70

= f0 rS [ 3(a +cos a sin a- 2 coss a sin a)- 8 cos a sin' a)] + fcpnrR' 'TT

k H Q

f rS [ (R)' J = - 0- l2(a +cos a sin a- 2 coss a sin a) - 32 cos a sin' a+ 24pn - , Equation 5.

96k r "

e

r = M0 + M.

(V0 + V0)r

fer' [12(a+cos asin a - 2 coss a sin a) - 32cosasin•a+24pn (~)'., 96k r

fer' (2 sin' a + 3 sin a cosz a- 3 pn 'TT cos a- 3 a cos a) 6k

_ 12 (a+ cos a sin a - 2 cos• a sin a)- 32 cos a sin' a+ 24 pn ( ~ )' 71'

16(2 sinS a+ 3 sin a cos2 a- 3 a cos a - 3 pn 'TT cos a , Equation 6.

CM+SM . . . - whereCM = 12 (a+ cos asm a-2 coss a am a) - 32 cos crams a,

16(CV -SV)

SM = 24 pn ( ~)' 'TT, CV = 2 sinS a + 3 sin a cos2 a - 3 a cos a

and SV = - 3 pn ., cos a

The observation is made that for any value of k or a, CM and CV can be computed. Table 1 has been computed for the values of k of .10 through 1.0 and the respective angles are noted.

Using these values, the curves on Figure 1 were constructed with k and a as ordinates and CM and CV as abscissas.

By the use of Equation 6, the neutral axis can be located. This is done by assuming various values for k until one is ob-

tained that approximates the~ of the external forces. Using r

the curves, this determination is rather easy to obtain.

The unit stress in the concrete is found by Equation 5;

f = 96Mk 0 (CM +SM )rS

Then, the unit stress in the reinforcement is found as f,

56

nf0 [R+r( l -2k)) . 7

2 kr , Equatlon

NOMENCLATURE

N = number of dowel bars A, = area of one bar in square inches

D = diameter of dowel bar circle in inches P = minimum total of vertical loads in pounds at the

juncture of pedestal and concrete slab (section A-A)

M = maximum bending moment in inch pounds at the bot­tom of pedestal (section A-A)

p = ratio of area of steel to area of concrete n = ratio of modulus of elasticity of steel to that of the

concrete r = radius of concrete pedestal in inches

R = radius of dowel bar circle in inches M = external moment at the section V

0 = total vertical force in the concrete

V, = total vertical force in the reinforcement (dowels) Me = resisting moment of the concrete M0 = resisting moment of the reinforcement

f0

= maximum unit stress in concrete in pounds per square inch

d = diameter of circular pedestal f, = maximum unit stress in the reinforcing steel in pounds

per square inch 2kr = the distance to the neutral axis measured along a

radius from the point of maximum stress in the con­crete. (kd)

2 a = the angle aubtended by radii drawn from each end of the chord which forms the neutral axis

e = eccentricity in inches of M/P ##

2,000,000

1,000,000

500,000 ...

~ ... c "g :::; 0 m g 0 0.. o;

~ .... .. ..... o; d-en ~ ......

~ ..0 ...J. .c

::t: .,. 1: ... ... ... ::t:

~ 10 ~

0 ...

2 .. 0

"" ~ ... c .E .. 20 .. m ..

t: ... 30-----:~ ... 40 40 ... .. .. 0 ..c

~8 ... cr

t2 1: 30 o; E .. .. ... E 100 0 "' .. 0 2 20 lil

-~ ... .. "' c t2 100 0 ·e 0 0 ...J

-= 90

·c; ... 10 -::; 80 en .. ::0

o. ""' 10 "" 0 c Q: 60 => 5,000 0 ::t: en

0 50 0

... _; 40 . 5

...J ... .30 u

c ~ .:! ... 0:

1,000 20

FIGURE 1-This nomograph calculates the overturning moment and the unit soil loading in tower foundation design.

Short Cuts to Tower Foundation Design

Graphic solutions to unit soil loading and loading caused by the overturning moment will speed up your foundation calculation time

J. F. Kuong Atlos Powder Compony Wilmington, Del.

"FOUNDATION DESIGN For Stacks and Towers" by V. 0. Marshall was the subject of a special supple­ment published in the May 1958 issue of PETROLEUM

REFINER. This article, in turn, supplements Marshall's article in that it presents two nomographs for short-cut meti10ds to the analytical formula techniques requiring trial·and·error calculations.

In designing foundations for self.supporting towers, with respect to the supporting soil, two main considera· tions are taken into account: a) the unit soil loading, and b) the tower stability. These two factors must be studied. The first so that the maximum load the soil supports will not be exceeded and the second to prevent overturning of the tower by external forces, such as those caused by the wind pressure acting on the tower.

Calculations for this type of foundation requires a trial-and·error procedure. The size of the foundation is assumed. Then, the soil loading and stability are

51

50

40

1,000,000 30

900,000 800,000

700,000 -600,000

G) 20 G)

u..

500,000 .!: G)

"' 400,000

0 CD -0

300,000 .... 10 Q)

"' - 9 "0 G)

c: E 8 ::J .~ 0 C) 7 0...

200,000 -c: .... 6 0

..<:: - en 1- 5 G)

~ "0 c: ... _J

0 4 Q)

~ (.)

c: 100,000

Q) .... 3 G)

0 90,000 -en Q)

80,000 a:: c:

0 70,000 a::

"0 0

60,000 2

0 ...J - 50,000 .&; 0'1

'Q; ~ 40,000 ~

30,000

20,000

10,000

Equation:

~ t Where :

_i, Con Be Either

S1 Or s, .. And ! Is W Or Wr, Respectively.

K, Is The Area Shape

Foetor. _. u..

d, Ia Tile Short cT en

Diameter Of The Sou .......... .d _J ---en "0 0 0 .... _J

0 - "0 (.) 0 0 u.. 1000 Square ~

G)

;-:a;soclogon 0 ~ 0

~·'" .... ,;; j ..<:: en 0 G) .... .7854 Circle E <t ::J

::.:: E c: ~ 0'1 c: "0 0 0 _J

'(5 en

c: ::::>

en

No111ogroph No. 2

Key : W -d -- R -K -s

100000 90000 80000

70000

60000

50000

• 40000

30000

20000

10000 9000 8000

7000

6000

5000

4000

3000

2000

1000 900 800

700

600

500

FIGURE 2-Use this nomograph to find the minimum and dead soil loading for tower foundation design.

checked and used a'! a criteria to determine the suit­ability of the foundation size originally assumed.

In estimating the maximum soil loading, two kinds of loading must be considered, namely: a) the unit soil loading due to the dead load (which includes the weigh t

58

of the empty tower, appurtenances and foundatio1., as well as the earth fill on top of the foundation base) and, b) the unit soil loading caused by the overturning moment produced by the wind or any other lateral forces acting on the tower.

The total soil loading is, therefore,

S = S1 + S0

where: S = Total soil loading, psf.

S\ = Unit soil loading, dead load, psf. S

0 = Unit soil loading, moment, psf.

S1 and So are calculated as follows:

when it is erected and empty and does not include aux­iliaries. As explained in more detail by Marshall, 1 in calculating the stability of the tower, the maximum soil loading must be used in equation (7) as defined in this paragraph. Calling the minimum soil loading S1m and W T the minimum deadload, we have:

s - WT (9) tm- K • d2 W M S1 =-

3- (2) and S0 = T

where:

<3> and the condition of a perfectly balanced system, as ex­plained in m:>re detail in the reference article, is

a, is the area of the base of the foundation sq. ft. M1, Overturning moment about the base of the foundation,

foot-pounas. W, is the weight of the empty tower plus the weiftht of

the foundation itself, including the e:~rth fill on top of the base (minimum dead load), plus the weight of auxilinries to include the weight of the tower contents and appurtenances, in pounds.

Z, is the Sl'Ction modulus of the base of the foundation which varies with the geometric shape, cu. ft.

Now, since a= K (d} 2 (4)

and, z = F (d) 3 (5)

where d is the short diameter of the foundation base and K and F are proportionality constants for a given geometrical shape of the foundation base (octagonal, round, etc.), and since for cylindrical towers the over­turning moment is given by

M1 = 0.0025(V)2 • D0

• H • L (6)

equations (2) and (3) can be written as follows,

Here,

w s1 = K. d2

S _ 0.0025(V)2 • D0 • H • L o F • ds

(7)

(8)

V, wind velocity, mph. D0 , diameter of tower including insulation, ft. H, height of tower, ft. L, lever arm of wmd load, in feet, calculated as follows:

L = h, + H/2. h1, height of foundation, ft.

Furthermore, the condition of poorest stability occurs when the tower is installed by itself. In other words,

About the Author

J. F. Kuong is a process engi­neer for Atlas Powder Co., Wil­mington, Del., where he works in process improvement, trouble shooting and cost reduction. He is currently in charge of a technical section doing technical-economic studies, process improvement work and technical support for line su­pervision. Holder of a B.S. degree in chemical engineering from the University of San Marcos, Peru,

and M.S. degree in chemical engineering from the University of Pennsylvania, he has been with Atlas since 1954. Kuong worked in the Technical Depart­ment, Atlas Point Plant until 1956, when he became technical assistant to the production superintendent. Kuong has been a process engineer since 1957.

s,m =so (10)

and for an actual system Stm should not be less than So.

Nomographs. Based on equations (7), (8} and (9), two nomographs have been prepared which reduce the time required in repeated trial-and-error calculations. The first, based on equation (8), gives directly the value of the unit soil loading due to the overturning moment, Mr, when V, D0, Hand h1 are known.

The second nomograph solves both equations (7) and (9) when W, WT and d are known. The nomograph is the same since ('quations (7) and (9) differ on ly in the value of W which is required to calculate S1 or S1m.

Example. Consider the same example given in the Marshall article (to which reference is made for de­tailed calculations) and compare the solutions obtained for S,, So and S1m using the nomographs presented here, with those given in the original reference.

The following data arc given: Tower diameter inc. insulation, Do = 4.5 ft. Weight of empty tower ........................ 30,000 lbs. Wtight of assumed concrete foundation volume

based on octagon-shaped base ................ 63,000 lbs. Weight of earth fill ............ . .............. 32,700 lbs. Minimum dead load, WT, (30,000 + 63,000 +

32,700) ... . ... . .......................... 125,700 lbs. Weil{ht of auxiliaries, insulation, platforms, piping,

etc., plus liquid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48,500 lbs. Total h('ight, I I, 54 ft. Height of foundation, hr, 6 ft. Assumed short diameter of octagon-shaped based, d, 13.5 ft. K, area proportionality constant for octagon base is 0.828. F, section modulus proportionality constant for octagon-shaped

base is 0.1016.

To calculate So, multiply Do • H = 4.5 • 54 = 243.8. Calculate L = 6 + 54/ 2 = 33. Enter 243.8 on DoH scale (Figure 1) and align with L = 33 on L scale to intersection with first reference line. With reference line as a pivot, align pivot point with V = 100 to obtain M 1 = 200,000. Connect 200,000 on Mr scale with reference point for octagon section modulus fac­tor and second reference line. Finally, align pivot point on second line with d = 13.5 and read So = 800 lbs./sq. ft. on extreme left scale. The calculated value given in the reference is 803.

To calculate S1 (use Figure 2). Align W = 174,200 on left scale on nomograph No. 2 with d = 13.5 and reference line. Connect pivot point on reference line with octagon area factor and read S1 = 1150 lbs.fsq. ft. The reference article gives S1 = 1155.

To calculate S,m (use Figure 2). Repeat procedure outlined just above except use WT = 125,700 instead of W = 174,200, and read S1m = 830. This value com­pares with 830 as given in the reference.

LITERATURE CITED 1 Manhall, V. 0., Foundation Design Handbook for Stocks and Towen,

PETROLEUM REPINEa, 37, No. 5, Supplement (1958).

59

FOUNDATIONS ...

. .. . ~ . .

... .

Foundation Design for 8-Legged Vessels

Using one general equation for bending moment, the reinforcement bars for the entire foundation can be calculated

Andrew A. Brown, Olefins Division, Union Carbide Corp., South Charleston, W. Va.

THE FOUNDATION for the 8-legged cylindrical vessel shown in Figure 1 can be designed with one general equation.

Because of the relatively low height of these vessels, the unit stresses at full load are usually not influenced by wind or seismic forces. That is, when the allowable unit stresses are increased by one-third for loading consisting of combined maximum vertical and horizontal forces the elements of the foundation are not usually overstr~ssed by such loading.

Base Slab. The base slab for the foundation is octagonal. Formwork for this shape is less costly than for a circular shape, and the distribution of stresses is more uniform than for a square. The base slab is assumed to be divided into four equal bands as shown in Figure 2. This is a view looking up from underneath the footing. The out­lines of the overlapping bands form soil pressure prisms. One of them is included in all four bands, six are in three, four in two, and two are in one band only.

Since all four bands are identically loaded, one will be removed and treated as an independent simple beam span. The reactions are the pier reactions. Section A-A (Figure 2) is formed by a plane passing through the center of the piers. The magnitude of the loads or soil pressures have been drawn to a vertical scale to show the fraction of the uniform load that is supported on the span. If the load prism is in all four bands, the load in­tensity is one-fourth of w. If the load prism is in three bands, the load intensity is one-third and so on.

If we let D equal the short diameter of the octagon, in feet, and P the total load or soil pressure on it (excluding the weight of the top fill and the concrete slab) the uni­form load becomes PI A = P /0.828D2 = w in pounds per square foot when P is in pounds.

The reaction for the beam is P /8. With these loads, the table of areas and moments is

Fig. 1-Elevation of typical vessel and foundation.

computed (Table I ) . The last column gives the moment of the respective load prisms about the center of the span. The values in the other columns are labeled and are self­explanatory. The total moment is - 0.0328wD2 for one band. The bending moment on a width of beam of one foot is

M O.l035wD: 0.0328wDa = 0.414D ( D/ 2 - x) - 0.414D

= [.25(D/2- x) - 0.0795D]wD

where "x" is the distance from the outer edge of the octa­gon to the center of the pier or reaction.

Since two-way reinforcement is to be employed, the influence of the two bands which cross this one at an angle of 45 degrees will have to be taken into account. A section of unity width is removed from the center of the span and the value of moments imposed at the center by these bands are shown in Figure 3.

Let m = moment on the main band acting on unit

61

FOUNDATION DESIGN FOR 8-LEGGED VESSELS

w:..f..: __ P __ A D2 X .828

f----SHORT DIAMETER OF OCTAGON,D,FT.-----1

DISTANCES TO C. G. t--------t-,43096 DI

r---D/2-x

r----D/2------1

t SPAN

SECTION A-A SHOWS LOAD DISTRIBUTION ON THE SANDS

Fig. 2-Base slab plan looking up underneath the footing.

TABLE 1-Table of Area a a nd Mome nts

Mom. Height About

Area of Soil Volume Dlst. to Center Sec. Prism Base Bear. Force Cent. Grav. Spnn

I. ....... . 042893D• "' .042893D2w .43096D .018485D'w

II ....... .07Hl67D• w/2 .0.155.11D2w .31904D .01133G6D2w

111-a .... .007:J59D• w/3 .0024531D2w .235700 .000578DJw -

III-b .... .01471902 w/3 .0040063D2w .1()667D .000817D2w

IV ..... .07Hl67D• w/4 .oJ7766D2w .09048D .001607702w

Totals .... .207!0402 R- .O:l282.1603to .10355D>w

Use .032SD2w

Note: Th~ valu~s in this table were extended further than th~ ~tr~ngth of th~ materials of construction and soil bearing determinations warrant. This was done to check the work. For Instance. the total area should equal .4142D X .6D, and tb~ total fore~• or reaction shoul<l equal one-eighth area of an octagon.

62

UNIT WIDTH

Fig. 3-Moments about section of unity width.

width. Then the width at a 45° angle is 0. 707 and the moment is 0. 707m. The total moment for the two bands is 2 (0.707m) = 1.4lm, but these are at 45 degrees to the main band. Accordingly, the component to be added is m(1.414 X 0.707) = m and the total moment is 2m.

For one foot width of beam, the bending moment at the center becomes M = 2[.25(D/2-x)-0.0795D]wD = [112(D/2-x) - 0.159D]wD (GeneralEquation)

With this moment, the required reinforcement can be computed. In sizing the bars one should keep in mind that the steel is located near the top of the slab and the permissible bond stress is 3.4 Vfc'/Bar Diameter but not to exceed 350 psi.

The pier reinforcement is computed by the usual for­mula used for column design. The minimum steel require­ment would normally govern. The maximum tensile stress in the pier rebars is obtained when the vessel is empty and maximum wind or seismic forces are imposed thereon.

The concrete is under maximum stress when the vessel is full and all other loads are applied.

If the octagon is much larger than the outside to outside distance of the piers, a section should be investigated at the plane of the outside pier edge. Reinforcement would then be required in the bottom of the slab. This projec­tion beyond the pier reduces the bending moment at the center of the span. # #

• •

1-

, ":.'1

About the author

ANDREW A. BROWN, Captain, Civil En­gineer Corps, U.S. Naval Reserve, is a Senior Engineer, Ole fins Division, Union Carbide Corp., So. Charleston, W. Va. Mr. Brown's professional e:r:perience in­cludes several years in the Bridge De­partnumt, State Road Commission of West Virginia, and he has served as a bridge consultant for several cities. During his 1! y ears of active duty in the U.S. Navy some of his billets were:

Public Works Officer, Naval Air Station, Hampton Roads, Va., Naval Air Station, Kaneohe Bay, Hawaii, and Naval Station, San· Juan, Puerto Rico; Design and Construction Officer, Fifth Naval District, Maintenance and Operations Officer, Eleventh Naval District, and Assistant Public Works Officer and Maintenance Superintendent, Naval Air Train­ing Basea, Pensacola, Fla. He is a member of International Association for Bridge and Structural Engineers, SAME, ASCE, and has BSCE and CE degrees from West Virginia University.

Pressure Vessel Foundation Design

For vertical pressure vessels, the old middle third rule

requires a safety factor of 3. These data show that a

factor of 1.5 is quite in order

J . A. A . Cummins Hudson Engineering Corp. Houston

1.9

l.S

z y

- I

IF THE FULLEST economy is to be realized in the design of a foundation, a complete understand­ing of its action under various load­ing is required. This is particularly true when the wind loading resultant falls outside the middle third of the foundation cross section and uplift occurs. In this case, the soil pres­sure, or pressure on the soil, varies in a different manner than when the resultant falls inside the middle third. It is emphasized that a factor of safety against overturning of 1.5 is quite in order, whereas the mid­dle third rule gave a factor of safety of 3. The use of this data will result in a precise, and hence more eco­nomical design and will be consist­ent with the safety factors derived from the American Standard Build­ing Code Requirements A58.1-1955, and the ACI Building Code.

.; •• ConJI ~ •• C!IHlll c:t=J ...

Advantages of Square Base. For vertical pressure vessel foundations, a square base is preferable to an octagonal or round base mainly be­

cause of the complications involved in laying out the steel. An octagonal base requires at least three layers of steel, one on top of the other, and consequently a greater depth of con­crete.

0.8 L_ __ ....L._ __ _._ __ ___J. ___ .~.-. __ ..J_ __ _._ __ ---i

1.5 2 .0 2 .5 3.0 3.5 4.0 4.5 5.0

.,..., B BW /YY "2e" 2M

FIGURE l-In Case 1 the resultant is outside the middle third.

63

The base area is practically the same in both cases, but the octagonal base uses considerably more steel. Furth­ermore, the form work for an octagonal base costs more than a square one of the same area. A square base also permits closer spacing of· columns than the octagonal base.

The base must be set below ground, the bottom being below frost line and on undisturbed soil of known char­acter. A pedestal is required to convey the load of the vessel to the footing. Ideally it should be circular, but an octagonal,pedestal is cheaper to construct. The mini­mum depth below grade of the top of the footing is often governed by the depth required to accommodate the various pipes which are necessary wherever this type of vessel occurs.

Overturning And Soli Pressure. In any design, the factor of safety against overturning and the maximum soil pressure must be computed first. The size of the base is then determined by trial and error. Figure 1 shows the relationship between the soil pressure p at the side of the footing caused by wind normal to axis YY, and the soil pressure p' at the corner caused by the wind nor-

mal to axis ZZ. The ratio~ varies with the eccentricity p

and hence with the factor of safety :e = '1· The curve

shows three conditions: Case I, when the resultant is outside the middle third and uplift occurs for bending about both axes; Case II, when e11 is less than

: , but there is still uplift at the corner when bending

about ZZ; and Case I II, when bending about both axes produces no uplift.

I t can be seen that about '1 = 2.8, p = p'. Below this value, p' is always less than p so that it is only necessary to investigate p. When '1 is greater than 2.8, the pressure on the corner is greater than that at the side. The curve is useful here to determine p' from the easily calculated value of p. It may be observed that the maximum ratio

of~ will be 0.~4 = 1.19. For cases where '1 is greate~ p W M X 6v2

than 3.6, it is easier to compute p' =B2plus B3

in the usual manner. It is usual when computing soil pressure to test for

dead load plus test load with water and no wind, or dead load plus operating load plus wind load. Some designers still use dead load plus test load plus wind load together, but this is not necessary since the vessel is usually filled with water only once or twice in its life for testing pur­poses. However, some companies advocate the filling of vessels with water when hurricane warnings are re­ceived, in which case the latter condition should be used for design, but this practice is not very common.

For computing the minimum factor of safety against overturning, the e1npty weight of the vessel should be used together with the weight of the soil and footing.

64

y M N z

0 b

B

z d c y M N

FIGURE 2-The octagonal pedestal is reduced to a square or equivalent area.

Allowable Soli Pressure. Allowable soil pressures quoted by most soil engineers are usually given as a figure in pounds per square foot at a certain depth. This is the allowable pressure in addition to the weight of soil already there. Thus, allowance can be made for the weight of original soil above foundation depth when computing maximum soil pressure. This has a net ef­fect of increasing the given allowable soil pressure by the weight of soil above the given depth. If this figure is used, allowance must be made for all backfill which may lie above the foundation as well as the weight of concrete and maximum weight of the vessel.

Wind Pressure. The American Standard Building Code Requirements for Minimum Design Loads in Buildings and Other Structures, A58.1-1955, is the culmination of much study of wind pressures in the United States and contains recommendations for design pressures for different areas. It shows how wrong it is to take an arbitrary wind pressure and apply it to all localities and to any height above ground as it fre­quently done today. Specification writers often specify a figure of pounds per square foot on a projected area which in many cases is too high for the majority of ves­sels, but which for very tall vessels, is actually too low. Also, they will sometimes specify a wind velocity and omit stating the height at which the velocity is to be taken. The Code gives design wind pressures recom­mended for any location in the U.S. and for any height above ground level. In addition, minimum recom­mended seismic factors are given and a schedule for recorded earthquakes in the U.S. with a map showing the locations of their epicenters.

Concrete Design. The concrete base should be de­signed according to the latest ACI Building Code ( ACI 318-56) . In Figure 2, the bending moment on the foot­ing is calculated on Section MM and the diagonal ten­sion on NN ~ith bending about YY axis. The octagonal pedestal is reduced to a square of equivalent area and the length of the side C' = 0.91 X dia. of an octagon. It can be shown that by turning the equivalent square through 45 degrees and investigating bending about the

ZZ axis, the moment is never more than y2 X MMM. Since the two-way footing will be designed for bending 'about YY, the components of the steel in the diagonal

direction will be 2 X l - = y2 X the area of the steel y2

in the YY direction. Consequently, bending about ZZ does not enter into the concrete design.

The ACI Code at present makes no difference for a square and rectangular footing in the location of the critical section for bending. It is suggested that a more correct estimate of moment for square footings would be obtained by taking moments of the loaded area beef about Section MM and distributing the steel computed over (ef plus 2d) where d is the effective depth of the base. This is particularly true with a small pedestal on a large base. The steel outside this critical section could then be placed at wider spacings. At present, the ACI Code specifies the moment to be taken and the steel to be distributed evenly over the full breadth of the base on Section MM. The diagonal tensile stress is to be computed from area bcgh on Section NN of width gh. The computation of bending moment and diagonal ten­sion about these sections is best handled by use of formulae. These must vary according to whether the conditions are Case I or Case II (see Figure 1) . Case III will be similar to Case II. The method of design is probably best illustrated by the following examples:

EXAMPLES A typical absorber is taken, for example, with founda­

tions to be located in ( 1) Nevada, (2) Oklahoma, (3) Central Texas (Figure 3) . The vessel is 84 inches ID X 99 feet-6 inches S-S. The allowable soil pressure is 4000 PSI at 5 feet below grade. The shell thickness is 3.5 inches. Using 4 inch insulation gives 99 inches or 8.25 feet OD.

Erected weight empty Weight of water to fill Maximum vessel weight Operating liqujd load

409,620 lbs. = w. 244,500 lbs. = w .. 654,120 lbs. = wm

Pedestal 9 feet-3 inches across flats; weight = 70.84 X 4 .25 X 150

Operating design load for mat

=WI

The following abbreviations have been used:

409,620 Jbs. t 28,800 Jbs. ~ 45,100 lbs.

483,520 lbs.

M,.r = Wind moment about bottom of footing at center, lb. ft.

T 99'-6"

l H=IOG'·O"

FIGURE 3-Example Figure. A typical absorber is studied with foundations located in Nevada, Oklahoma and Central Texas.

M. = Earthquake moment bottom of footing, lb. ft. C' =Side of square of equivalent area to octagon, ft. s, = Soil pressure for mat design, PSF S0 = Soil pressure due to total operating load, PSF St = Soil pressure due to total test load, PSF S,. = Soil pressure due to wind or earthquake, PSF D = Depth of concrete, ft. d = Depth of steel in concrete, inches B =Side of square base, ft.

']min= Minjroum FOS against overturning ']o = Operating FOS against overturning e0 =Operating eccentricity, ft.

V 0 = Shear at Section N, lbs. V m = Shear at Section M, lb/ft Mm = Moment in footing on Section M, lb ft/ft Sa= Allowable stresses in accordance with ACI 'Code, psi

Mu = Moment in footing due to uplift at Section M, lb ft/ ft

~0 = Sum of bar perimeters in one foot width, inches s. = Shear stress, psi

f = Depth of soil above base, ft.

W.plusW1 = 438,420

Notes. ( 1) Unit weight of soil in calculations has been taken at 90 lbf cu ft, being the probable minimum weight if unconsolidated.

(2) For wind moment calculation allow 2 feet width for ladders, pipe, platform, etc., i.e.

65

total OD vessel = 8.25 plus 2 = 10.25 feet. This 2 feet is conservative and may be re­duced by an analysis of all the extraneous projections.

Example (I)-Nevada. A58-1-1955 gives a 20 PSF pressure zone. Wind pressure 0- SO feet= 15 X shape factor 0.6 = 9 PSF en vert. proj.

SO- 50 feet= 20 X shape factor 0.6 = 12 PSF

50- 100 feet= 25 X shape factor 0.6 = 15 PSF

100 feet up =SO X shape factor 0.6 = 18 PSF

M,.r = 10.25· ( 18 X 6 X 108 plus 15 X 50 X 80 plus 12 X 20 X 4-5 plus 9 X SO ,X 20)

= 10.25 ( 11,664 plus 60,000 plua 10,800 plus 5,400) = 900,6061b/ft

Let B = 19 feet-6 inches and D= 1.75feet

Weight of base 19.52 X 1.75 X 150 Weight of soil above base

90X (280-71) 3.25 Operating load on soil Operating liquid load -deduct Minimum direct soil load Water to fill Max. test load on soil Soil pressure under test load

W 0 = 483,520 lbs. wb = 99,820 lbs.

W 1 = 90,380 lbs. W

0- 673,720 lbs.

W 1 = -28,800 lbs.

W min= 644,920 lbs. W,. = 244,500 lbs. wt = 889,4-20 lbs.

2,341 PSF

As Nevada is a region of recorded earthquakes, see A58.1-1955, a seismic factor of 0.1 is taken.

M. = 0.1 (438,420 X 58.5 plus 144,920 X 2.5) = 2,564,750 plus 362,300 = 2,927,050 lb. ft/ft.

This moment must be used since it is greater than M.,.r.

_ WmlnXB '1mln- 2 M•

644,920 X 19.5 = 2 15 1.

5 2 X 2,927,050 . >

W 0 X B = 673,720 X 19.5 = 2.24 '1o = 2M, 2 X 2,927,050

M e =-•-

o wo

- 2,927,050 - 673,720

= 4.35 feet

From curve, '7o < 2.8 and ;, > 1 i.e. max soil pressure at side

4- wo Soil pressure =-=-=-':-:-

3B (B-2e) 4- X 673,720 = 4-,265 PSF 58.5 X 10.8

Subtract wt. of original soil at given depth 90 X 5 = 4-50 Max. soil pressure 3,815 PSF < 4,000

Concrete design for 2,500 psi at 28 days f

0 = 1,125 f, = 20,000

adding 33y; percent overstress for ACI 603(c)

f0 = 1,500 f, = 26,667

.iFor D = 1.75; let d = 17 inches; C' = 0.91 X 9.25 = 8.42

;For '10

< 3, proceed as follows:

X= 1.5 (B-2.) = l.5 X 10.8 = 16.2

Y= ~ (B-C' -d/6) = ~ (19.5-8.4-2 -2.83)=4.13

66

2 W "Y[ J V 0 = B X 2 ~ (C' plus d/6) (2X- Y) plus Y/3 (3x- y)

2 X 483,520 X 4-.13 19.5X (16.2)2

[~ .(8.42 plus 2.83) (32.40- 4.13) plus 1.28 (48.60- 4.13) J = 780 ( 159.08 plus 56.92) = 168,480 lbs.

v S = Shear Stress = n • ( 12"0' plus 2 d) jd

168,480 ....,..,~-=--=-:-7-:~=-:-~=- = 85 PSI < 100

( 101 plus 34) 0.867 X 17 u 2 W" B-C'

M 01 = 3B,Z

2 ( 2X plus u) where u = --

2-

u = ~ (19.5-8.42) =5.54

(5.54 )2 483,520 Mm = 58.

5X (16.2) 2 (32.40plus5.54) =36,700lb.ft/ft

WP = 45,100 lbs. 99,820 lbs.

wb + WP 144,920 lbs.

Steel in bottom 12 M01 = 12 X 36,700 s. X jd 26,667 X 0.867 X 17

= 1.12 in.2

Use No. 7 @ 6 inches on centers

KF = 260 X 12 X 172 = 75,300 > M - no compression steel 12

4-Mm check bond: V m ::;: (B _ C')

4 X 36,700 11.08

13,250 lbs./ft

- vm - 13,250 bond stress- jd l:o - 0.867 X 17 X S.S = 163 PSI < 267

The formula for uplift is approximate and applies only when BplusC'

x> 2

Mu= - (90fplus 150D) ---1 B2 ( 3 )2 8 '~min

- 380 (90 X 3.25 plus 150 X 1.75) 8 (0.396)2

= 4,135 lb. ft/ft

. 4,135 X 12 _ 0 13

. 2

Steel 10 top 26,667 X 0.867 X 17 - · 10

Use No.4@ 12 inches on centers

KF > M I• no compression steel

Vu = (90 f plus 150 D) (B- x) = 555 X 3.3 = 1,850 lbs./ft.

bond stress 1 850 .

0.867 X 17 X 1.6 = 79 pslok

Example (2)-0klahoma. A58-1-1955 gives a 40 PSF pressure zone.

Wind pressure 0- 30 feet 30 X shape factor 0.6 = 18 PSF on vert. proj.

30- 50

50-100

100up

40 X shape factor 0.6 = 24 PSF

50 .X 1hape factor 0.6 = 30 PSF

60 X shape factor 0.6 = 36 PSF

M ... r= 10.25 (36 X 6 X 108 plus 30 X 50 X 80 plus 24- X 20 X 45 plus 18 X 30 X 20)

= 10.25 (23\328 plus 120,000 phu 21,600 phu 10,800) =1,801,200 b. ft.

For 8 = 18 f~et; D = 1.5; f = 3.5 W, plus W 1 = 438,420 lbs.

WP = 70.84 X 4.5 X 150 = 47,820 lbs. ..;- 82 = S

1 = 1,501 PSF

Weight of base 182 X 1.5 X 150

Weight of soil above 90X3.5 (324-71)

Operating load on soil

Operating liquid load deduct

Minimum direct soil load

Water to fill

Max. test load on soil

Soil pressure under test load

Seismic factor 0.025 gives M.

610,010 X 18 '~miD= 2 X 1,801,200 3.05

W 0 " = 486,240 lbs.

Wb= 72,900 lbs.

W8

= 79,700 lbs.

w 0 = 638,840 lbs.

W 1 = -28,800 lbs.

W min= 610,040 lbs.

W w = 244,500 lbs.

W t = 854,540 lbs.

Wt/82 = 2,637 PSF

;;;;:: 652,000 lb/ft < M,.r

638,840 X 18 2 X 1,801,200

3.19 and e0

1,801,200

638,840

= 2.82 feet

Consult curve for.-!:. for '10

= 3.19 and obtain.-!:.= 0.914 p' p'

. 4W0 Sot) pressure = 38 (8 _ 2eo) • P _ 4 X 638,840 0 p;-- 54 (18-5.64)

1 X

0_914

= 4,189 PSF

Subtract weight of original soil at given depth 90 X 5 = - 450 (at comer) Max. soil pressure - 3,739

< 4,000PSF

Concrete design for 2,500 psi concrete @ 28 days f0 = 1,125 f, = 20,000 and adding 33 ~ percent overstress for ACI 603 (c)

fe = 1,500 f, = 26,667

C' = 0.91 X 9.25 = 8.42 feet B = 18 D = 1.5 d = 14 inches

K ·= 1/B (C' plus d/6) = 1/18 (8.42 plus 2.33) = 0.6

S = M ... r X 6 1,801,200 X 6 = 1,853 PSF w 8 1 (18)a

B2 Max. shear V D = 12c 1 - K) [3 S1 (K plus 1) plus 2 s ... (K2

plus K plus 1)]

Note: if s ... < Ys S1

, then let S,. = 0 and allowable shear stress S, = 75 PSI

if S,. > Ys S1, compute V n as above and S, = 100 PSI

182 Vn=12(0.4)

(3 X 1,501 ( 1.6) plus 2 X 1,853 (0.36 plus 0.6 plus 1)] = 10.8 (7,~05 plus 7,264= 156,265lbs.

Shear stress S, (12 C' plus 2d) X 0.867 d

156•265

= 99 9 PSI< 100 (101 plus 28) X 0.867 X 14 ·

Design moment Mm = ~ (8- C')2 plus~ 8 24-B

(2 BS- 3 BZ C' plus C'S)

Note: if s ... < Ys s., then letS,.= 0 and allowable bending stress Sb = 20,000 PSI

if S,.. > Ys S1, compute Mm as above and Sb = 26,667 PSI

1,501 1,853 Mm =-

8-(18 - 8.42)2 plus X (11,664-

24 18 8,175 plus 595)

= 17,256 plus 17,518 = 34,774lb. ft/ft

Steel in bottom = Mm "{' 12 34

•774

X 12

Sb X .867 d 26,667 X 0.867 X 14 = 1.29 in2/ft.

Use No.8@ 7 inches on centers

4 M10

4 X 34,774- . V m + B _ C' =

18 _

8_41

:;;;;: 14,490 lbs./ft l:0 ·= 5.4 mches

KF = 260

X 12 X J..!.!L: = 50,960 > M, i.e. no compression 12 steel

v Bond stress = m

0.867 dl:0

14,490 0.867 X 14 X 5.4

= 221 PSI< 267

Note: if s ... had been < Ys S1, allowable bond stress = 200 PSI

Uplift-Since 7JmtD > 3, there can be no uplift. I t is usual to provide nominal No. 4 @ 12 inch centers in such cases in the top of the mat, both ways.

Example (3)-Central Texas. A58-1-1955 gives a 25 PSF pressure zone.

Wind Pressure: 0- 30

30- 50

50-100

100up

20 X shape factor 0.6 = 12 PSF

25 X shape factor 0.6 = 15 PSF

30 X shape factor 0.6 = 18 PSF

40 X shape factor 0.6 = 24 PSF

M,.r = 10.25 (24 X 6 X 108 + 18 X 50 ·X 80+ 15 X 20 X 45 + 12 X 30 X 20)

= 10.25 (15,552 + 72,000 + 13,500 + 7,200) = 1,109,580 lb. ft.

For B = 16 feet-6 inch D= 1.5 f=3.5

W0,. =486,240 ..;- B2 = S, = 1,786 PSF Weight of base

16.52 X 1.5 X 150 wb = 61,260 lbs. Weight of soil above

90X3.5(272-71) W,=63,320lbs.

Operating load on soil W0 = 610,820 + B2 = S0 = 2,244 PSF Operating liquid load-

deduct - 28,800 lbs.

Minimum direct soil load W miD= 582,020 lbs.

Water to fill W., = 244,500 lbs.

Max. test load on soil wt = 826,520 ..;- B2 = st = 3,036 PSF

67

Seismic factor 0.025 gives M8 = 651,960 < M,.t

Wmla X B 582.020 X 16.5 '~min= 2 M = 2 ·X 1 109 580 = 4·33 > 1.5

wt ' ,

610,820 X 16.5 17o = 2 X 1·,109,50 = 4·54

A! '10 > 3.6, proceed as follows:

M,.t X 6 S,.= BS

1,109,580 X 6 = 1,482 PSF (16.5)8

Soil pressure (on comer) = S0 + 1.414 S,. = 2,244 + 2,096 = 4,340 PSF

Subtract weight of original soil at given - 450 depth 90 ·X 5 - 3,890 PSF

Max. soil pressure < 4,000 PSF

Concrete design for 2,500 psi concrete at 28 Days.

"'o > 3, procedure similar to example (2).

Dowels. These must be provided to transfer any ten­sile force from the pedestal into the base. This force is transferred by bond from that part of the anchor bolts actually embedded in the pedestal. However it is usual to design dowels for the full tensile force whether the bolts go into the mat or not.

Let N = number of dowels in a circle Dd inches in diameter

Let M'' = the wind or earthquake moment at bot­tom of pedestal

Let A = area of each dowel; WR = minimum resist­ing weight

(48/Dd) M•-WR Then A= SA X N

About the Author

J. A. A. Cummins is a civil engineer working in design and construction for H\ldson Engi­neering Corp., Houston. He at-tended Nautical College, Pang­bourne, Epgland, for four. years;­then went to Royal College of Science and Technology, Gla.~­gow, Scotland, for four years tl study civil en'gineering. He started his career with a consulting engi­neering firm in Scotland in 1947. From 1951-54 he was ·a concrete structural engineer in England, and from 1954-56 was an engi­neer and superintendent on a project to construct a dam in Scotland. He joined Hudson 2~ years ago in Ontarior and has been with the Houston office for a year. A registered professional engineer, Cummins is a member of several technical societies.

68

For example (1), M" = 0.1 (438,420 X 56.75 + 45,100 X 2.25) =

2,600,470 lb. ft.

WR = cw. + Wp + WL) = 483,520

_ (48/104) X2,600,470 - 483,520_ 716,700 A - 26,667 X 32 - 26,667 X 32

= 0.84 in2

i.e., provide 2 at No. 9 and 2 at No. 8 at each of 8 pedestal faces.

(H/ 2) + L-D 56.5 For example (2), M" = (H/ 2 + L Mwt = 5s Mwt

For cases where the depth of the base D is small compared to overall height of tower, M'' can be taken as Mwf

WR = (W. + Wp) = 457,440

- (48/104) 1,801,200 - 457,440 A - 26,667 XN

373,890 26,667 X 32

= 0.44in•

).e., provide 4 at No. 6 at each pedestal face

For example (3),

_ (48/104X 1,to9,580 - 457,440 _0 3

. 2 A- 26,667 :X 16 - .I m

i.e., provide 2 at No. 4 at each pedestal face

Anchor Bolts. These may be computed as follows:

Let Db = diameter of bolt circle in inches.

N = number of bolts

M' = wind moment at base plate

W R = minimum resisting weight

(48/Db) M' - Wa Then Root Area A = f• X N

W R is usually the empty weight W •• but when the earth­quake moment is used, W R is the operating weight W • + W L

For example (1}, Using SAE 4140 bolts with allowable stress 30,000 psi

M' = 438,420 X 52.5 = 2,301,700 lb. ft.

A _ (48/104) 2,301 ,700-438,420 623,900 . 600,000 = 1.04 m2 - 30,000 X 20

Use 20- 1%-inch dia. bolts SAE 4140

For example (2),

, (H/ 2) 53 M ~ (H/2) + LMwr =ssX 1,801,200 = 1,646,000 lb. ft.

A (48/104) 1,646,000 - 409,620 = 30,000 X 16

Use 16@ 11,4-inch dia SAE 4140

350,070 480,000 0.73 inZ

For example (3) M' ~ ~;X 1,109,580 = 1,013,930

(48/104) 1,013,930-409,620 58,350 041. 2

A = 12,000 X 12 144,000 = · 10

Use 12 @ 1 inch dia. carbon steel. (Ys" dia would do here, but normally less than inch would not be used) ##

NOTES

69

~ "··

COMPUTER FOUNDATION ·:~···. ....

OESIGN . . f

... .

How to Calculate Footing

Soil Bearing by Computer

Here's an effective method for finding the maximum soil bearing under eccentrically loaded rectangular footings, programed for a small computer

Eli Czerniak, The Fluor Corporation, Ltd., Los Angeles

MOST OF THE STRUCTURES used in hydrocarbon proc­essing are, to some extent, affected by overturning forces which, like the vertical loads, must ultimately be resisted at the ground. The function of their footings, then, is to provide that resistance; so that all the loads-vertical, lateral, and overturning moments-can be adequately supported, without exceeding the safe bearing capacity of the soil.

The factors and causes contributing to the over­turning effects arc varied. Gusty wind pressures on exposed structures rising high above the ground is one; the seismic forces for the plants and refineries which are located within areas subject to earthquake shocks is another. Impact, vibration, crane runway horizontal forces, unbalanced pull of cables, sliding of pipes over supports, thermal expansion (or partial restraint) of horizontal vessels and heat exchangers, reactions from anchors and directional guides, eccentric location of equipment are some of the additional reasons for the lateral force design.

The actual mechanics for determining the maximum soil bearing under a footing are, of course, independent of any of the causes for the separate force components used in the various loading combinations met in design. T he computations arc the same whether the resulting overturning moment is from vertical loads which are located off-center (load time!\ eccentricity); from lateral forces that are applied at a given height above the foot­ing (fo1 ce times distance to bottom of footing); or by some combination thereof. Therefore, the techniques for tlw computational analysis, described in this article

FIGURE !-Computer-designed footings for a refinery.

will simply be based on the three resultants P, H, and M, for the vertical loads, horizontal forces and over­turning moments, respectively; applied at the footing centroid-without giving any special consideration as to how this combination of forces and moments was ob­tained. It should be mentioned, however, that when pro­portioning footing sizes in the design engineering office, the actual make-up of the critical load-moment combi­nations could be of economic significance. Figure 1 shows several computer-designed footings in a refinery under construction. As with the other engineering mate­rials, some increase is the allowable soil bearing is certainly justified when designing the footings for dead, live and operating loads, combined with the temporary lateral forces and moments. And due care must be exercised in establishing the proper design values. Ob­viously, no increase in allowable soil bearing would be advisable when the moments, about the footing center­lines, are due to the eccentricities of long-duration vertical loads. It should apply only to such loading

71

CALCULATE FOOTING SOIL BEARING . . .

FIGURE 2--Spread footing during construction.

combinations which are definitely known to include overturning effects of a temporary nature. Building codes, recognizing the improbability of the absolute maximums occuring simultaneously, usually permit foot­ings subjected to wind or earthquake combined with other loads, to be proportioned for soil pressures 33 Y3 percent greater than those specified for dead, live and operating loads only, provided that the area of foot­ings thus obtained is not less than required to satisfy the combination of dead load, live load, operating weights, and impact (if any).

Design Practice. With the almost infinite variety of soils encountered, the problem of determining the actual soil pressure under footings could be, to say the least, extremely complex. As foundation engineers well know, the distribution of loads and moments-on the footing, to the supporting earth beneath, is rather highly uncer­tain. Simplifying assumptions, however, come to the aid. According to current structural engineering practice, the soil bearing under the loaded footing is calculated from static equilibrium, and on the basis of the simpli­fying assumption that the footing slab is absolutely rigid and it is freely supported on elastically isotropic masses. From this follows a linear distribution of soil pressure against the footing bottom. For only concentric loads, then, the upward pressure is considered to be uniformly distributed over the full area of the footing, and hence

equal to ~ . When moment is also present, its contribu­

tion can be evaluated from the simple flexure formula

~e , provided that the resultant eccentricity e (com-

puted from ~) falls within the kern of the footing

area. By superposition, the maximum and minimum pressures are simply the algebraic additions of the direct

db . P Me d P Me an endmg components, A+ -1- an A- -

1-, respec-

tively. In order to obtain the net increase in pressures, the weight (per •tnit area) of the displaced earth and backfill should be deducted from the gross values. In

72

designing the concrete and reinforcing steel in the foot­ing, only the net pressures need be considered. When the position of the resultant eccentric load is outside the kern, straight forward superposition is not applicable because the pressure reversal implied by the flexure formula cannot occur in a footing on soil. When the overturning effects exist about two axes, the analytical confusion is further compounded. The technique de­scribed in this article, however, is completely general, and hence effective for all cases, with resultant load locations inside and outside of the kerns. A close-up of a spread footing during construction is seen in Figure 2. Under the superimposed loads, the upward soil pressure tends to deflect the projecting portions of the footing, until it would assume a slightly convex shape. The reader need not have any qualms about the previously con­jectured, absolute footing rigidity. As stated before, that assumption of perfect rigidity was made only for the purpose of facilitating soil pressure computations. This purpose having been satisfactorily achieved, the engineer must then tackle his next item on the agenda-the structural design of the footing itself. To accomplish that, he expediently relaxes the rigidity restriction, and permits the soil pressure against footing bottom to deflect upward (not too much though) the outer por­tions of the footing. To resist them, steel bars are added to compensate for the inherent tension deficiency of plain CuHClcL..:. L, isolated footings, the tensile rein­forcement is placed in two directions, (as can be seen in Figure 2) with the bars in one direction resting di­rectly on top of those in the other direction.

Biaxial Eccentricity. When the overturning moments are about two axes, the footing obviously, will bear most heavily on one corner, and least on the corner diagonally opposite. As long as the eccentricities from the resultant load-moment combination are sufficiently small to remain within the kern, the entire footing is under compression, and corner pressure can be computed from the well known formula

...:._ + M.c. + Mye1

A - IX -4 However, as the eccentricities increase and fal! outside the kern, the computations become quite complex, even with the simplifying assumption of the straightline pressure distribution. Because tensile resistance of soil sticking to the footing obviously cannot be depended upon, common practice is to ignore from the analysis that portion of the footing area over which the soil pressure would have been negative. It is the difficulty in determining the shape and size of the remaining "effective" portion which constitutes the major stumbling blocks in the efforts to achieve a mathematical solution. Depending on the location of the resultant of the applied loads, the effective portions of rectangular footings could well vary from a triangle, through trapezoid, to a full rectangle. The line of zero pressure (neutral-axis) establishes the boundary of what is to be considered as the effective footing area. From statics, the value of the resultant of the applied loads P must equal the total

I: D :j D ~ 4 4

~I..-

-i

y

y T

Q' ~T' .D

T .a T .a

1 j_l x 1 j1v X l X

o~ a ---1 o~a ~ ol-- a -----J ol- a -----l 1- a

FIGURE 3--Depending on the load location, the effective area can be one of five possible shapes.

reaction of all the soil pressure against the footing, and the location of P must also coincide with the line of action of that total soil reaction, which is at the center of gravity of the soil pressure prism.

For any known or assumed position of the neutra l­axis, the maximum soil pressure under the footing

corner equals the resultant load P divided by A - C!or -a

~x , where A is the effective footing area; Q0 • and Q07

are the first moments of area A about the x- and y- axes; a and b are the intercepts of the neutral-axis line on the x- and y- axes, respectively. The origin of the rectangular coordinates is taken at the footing comer where the soil pressure is maximum. Depending on the location of the resultant load P (in the quadrant of the footing with the corner as origin) the effective area can be one of five possible shapes. The load loca­tions that correspond to these shapes (with matching cross-hatch regions) are shown mapped in Figure 3.

Stability. For the resultant load P to be within the kern, the sum of the eccentricity-ratios in the x- and y- directions must be equal to or less than one-sixth, . E,. E1 1 t.e., 0 + T ~ 6· The footing is then fully under

pressure, and hence the whole area of the rectangle is deemed effective in the analysis. The intensity of the maximum pressure (at the corner) varies from an aver-

age pressure ~ when the load is located right at

the footing centroid (zero moment), to twice the average, when the load is at the edge of the kern. As the sum of the eccentricity-ratios increase to more than a sixth, part of the footing area becomes ineffective in the analy­sis: stability diminishes and the maximum soil pressure increases to mere than twice the average. Theoretically, the maximum soil pressure would approach infinity, and the stability zero, when the location of the resultant load P is placed along any of the footing sides. Though the abutting power of the soil might offer additional re­sistance to prevent actual overturning, its value is rather hard to ascertain. Common engineering practice is to neglect this contribution of passive pressure (except for very deep foundations) in the computations of either maximum soil bearing or stability ratio. The footings should be so proportioned, that there is an adequate factor of safety against overturning without a depend­ence upon lateral soil resistance; with a value of 1.5 being the minimum recommended. The weight of earth superimposed over the footing should be included in the stability calculations. Regarding the resistance to sliding,

73

0 1-

8 COMPUTE MAX. SOIL BEARING:

GIVEN P p = 0 A _ OoY _ Oox

a b

COMPUTE

p, =Po[ 1- ~]

PRINT: SOIL BEARING AT FOOTING

CORNERS

COMPUTE GROSS AREA: 1-------, A6 = DXT

COMPUTE: k , = IXY- Y,.Ooy k 2 = lxY- X.. Oox k 3 = lox- Y,.Oox k4 = loy- X.. Oov ks = Oox- YpA k 6 = Oov- X.. A

PRINTOUT: PROPERTIES,

PARAMETERS, % OF GROSS,

CYCLE

COMPARE PARAMETER b

TO OLD b

PRINT HEADINGS

AND GIVEN INPUT DATA

COMPUTE LOAD COORDINATES: X,. = D/ 2 - Ex Y,. = T/ 2 - Ey

COMPUTE INITIAL NEUTRAL-AXIS PARAMETERS:

a = 2Dl 100% b = 2T \BEARING

r- coMPuTe me - I I GEOMETRIC PROPERTIES I I OF EFFECTIVE* AREA:

L ~:_ ~x~ov,~o~l~, ~Y J COMPUTE PERCENT

FOOTING AREA UNDER BEARING

ADD ONE TO CYCLE 1------' COUNTER

STORE THE N.A. PARAMETERS USED

IN COMPUTING THE EFFECTIVE PROPERTIES:

OLD a = a OLD b = b

• See Fisur• 5 of "Concrete Support Analysis by Computer," Hydrocarbon Processing & Petroleum Rrfiner, Vol. 42, No. 8, 1963.

FIGURE -4-Logic pattern for computer program.

74

Exa mple 1

SIOL 0

15.000

SIDE l

1 0 .500

SOIL BEARING ANALYSIS OF RECTANGULAR FOOTI NG

AR.EA OF FOOT l NG

1!>7.500 SO.FT.

PROPER TIES

GIVEN LOAO P

250.000 KIPS

01' EFFECTlVE

11/fCCENTR JC IT J E S )( y

1. 000 l.OOO FT.

FOO TI NG AREA CYCLE ARt A :ji GROSS Ci Q I 1 I PARAMfTE RS

• 0)( C!Y ox OY H S TAitT WITH

1 157. 500 10 C.OCO fl26 . t!75 1181.250 5788. 125 ll$12.500 6201 .56 2 2 157.500 1CO . OCO 826 . 875 1181 . 250 5 788 .125 11812.500 6201.562

~ Pf. 0 SG IL BEARI~G AT •CORNER 3129 . 14 6 PSF

iil PT . l 1317. 310 PSF

• Example 2

SICE. C SIDE T AREA Cf FOCT ING GIVEN LOAD p

1!>.00 0 1.0.500 157.soo so.n. 100 . 000 KIPS

PROPI:RTHS OF EFFECTIVE

i PT. 2 50. 066 PSF

i PT . 3 l tl61.8H PSF

WIECCENTRICI TIES )( y

3.750 3. 250 FT .

FOOTING AREA

A 6 30.000 21.000 36.964 111 .1 13 36. 964 u . 1 n

C.YCl E Ai~EA t GRGSS ' Q I I I PARAMET ERS ())( OY ox OV XV A B

START WITH 30. 000 21.000 1 157.500 100 . 000 H26.615 1181.250 '.5788.125 11812. 500 6201.562 2 1.786 12.317 l 118. 231 75.067 501.779 11-6.501 2999.846 6696.222 2688. 006 17. 422 10.1"+8 3 86.690 , 5. 0 41 298.221 486.357 1516.686 404"+.463 1289.866 1 5~663 8.798 4 6&.718 't3.668 202.0'o9 357 . 855 888.882 2788.659 191.004 15.094 8. 147 5 61 . ... 83 39.036 166.973 309 . 317 680 . 1"68 2334.152 630 . 075 15 . 003 8 . 006 6 60 . (1 5.7 38. lJ 1 160.272 300 . 345 641.569 2253.038 601.140 15.000 a.o oo 7 60 . 000 38.095 l6o . ooo· 300 . 000 640.000 2250 . 000 600.000 15 . 000 8.000

@ PT . 0 a PT. 1 i PT. 2 i PT . J SOIL BEARI~G AT CORN ER 5000.000 PSF PSF PSF . 000 PSf

FIGURE 6-Computer printout of Examples 1 and 2.

common practice is to assume that it would be provided through the friction developed at the footing bottom.

Computer Program. The most difficult part of the problem (in both manual design and in formulating the procedures for sequential electronics computation) is determining the position of the neutral·axis which is taken as the boundary line cf the c!Tective footing area. TllC basic computer routine developed for solving biaxial eccentricity problems in reinforced-concrete, described in a previous article/ can also be used to solve footing soil bearing problems. T hat program was modified, so that title headings and data in the printed results wou ld comply with the usual nomenclature applicable to foot­ings. The formula~ for the neutral-axis and the effective section properties arc the same as given in the previous article, and therefore will not be repeated here. For background and development of the formulas and criteria the reader is also referred to the writer's paper "Ana­lytical Approach to Biaxial Ecc(:ntricity. " 2 T he logic pattern used in formulating this program for the small computer is shown flow-charted in Figure 4. From start to finish, load ing of the program deck and data cards, computations and the print-out of results for the two examples cited, took less than one quarter of a minute.

Example 1. The plan of a footing used in a rigid frame structure supporting several heat-exchangers is

shown in Figure 5. Determine the maximum soil-bearing for a total vertical load of 250 kips (weight of concrete foundation and earth backfill included) , located eccen­trically with respect to the footing centerlines, at a distance of 1'-0" from each centerline.

Solution.

I

-(D

!Ey p 1-~+

I I

-0

1 .

Ex

P = 250kips E = 1'·0" X

E = 1'-0" y

t

'

.

o, .. ..... ·-- ---- 15 I- 0 II

FIGURE 5-Example L

2

t . ....

75

CALCULATE FOOTING SOIL BEARING

---,..._.,t.~--1"1•¢-.. ;;;.;------

FIGURE ?- Distribution of soil bearing under footing.

For these eccentricities, the resultant load is obviously within the footing kern. Hence, the full footing area is under bearing, and the position of the neutral-axis line, falling outside the footing, has no effect on the geometric properties used in subsequent calculations. And since the computer program was set up to start with the full rec­tangle (by using neutral-axis parameters equal to twice

About the Author Eli Czerniak is a principal design engineer with The

Fluor Corp., Los Angeles. He coordinates computer applications for the Design Engi­neering Dept., reviews manual tech­niques and develops new methods and procedures better adaptable to systems conversion in automating the design and drafting of refinery units. Mr. Czerniak received a B.S. in engineering from Columbia Uni­versity in 1949 and an M.S. in Civil Engineering from Columbia in 1950. He is a registered engineer in Cali­fornia and has published a number of technical articles. He has had field experience as a civil engineer

Czerniak and worked in design and drafting with Arthur G. McKee Co. in Union,

N. J., for two years before joining Fluor in 1953 as a structural designer. He soon headed up the structural design and drafting on various projects until assuming his present position.

76

footing dimensions, see Figure 4) , the correct location o{ the neutral-axis line is obtained within the first cycle. Note: The test for convergence requires that the neutral­axis parameters remain the same throughout two con­secutive cycles, and whence the extra cycle shown in the computer printout (Figure 6) results.

Having determined the position of the neutral-axis, the computer next calculates and prints the maximum soil bearing (at corner used as origin), as well as the bearing at the other corners. The soil bearing diagram shown in Figure 7 helps visualize the results.

Example 2 . What is the maximum soil bearing, if because of additional overturning effects on the structure, the load specified for example 1 is reduced by 150 kips of uplift, while the eccentricities are increased by 2'-9" and 2'-3", in the x- and y- directions, respectively. Draw separate diagrams of the soil bearing under the footing for both examples.

Solution. Resultant P = 250- 150 = 100 kips

. . . ~Ex= 1.00 + 2.75 = 3.75 ft. Eccentncttles 1 Ey = 1.00 + 2.25 = 3.25 ft.

Stability t S.R. = 7.50/3.75 = 2.0 Ratios 5 S.Ry = 5.25/3.25 = 1.62 > 1.5

The load is now outside the kern, and with the large eccentricities used in this example, stability against over­turning could be critical and should be checked first. It is conservative to investigate the stability for each of the two directions separately, since in rectangular footings stability in any diagonal direction lies in between the two rectangular components. These two component values are shown in the readers' interest. The overall stability· ratio for the diagonal direction was also computed, and found to equal 1.84. Now, with the resultant load being outside the kern, part of the footing area must therefore be neglected. Noting that the eccentricity in the: x-

direction equals ~ , it is apparent that the location

of the load is on the dividing line between types I and III (see Figure 3). The limit of type I effective area is reached when parameter a becomes equal to dimension D, (and at which point type III begins). By observation, then, parameter a is known to be equal to 15.0 feet Such deduction would, of course, be helpful in reducing the volume of computations when attempting manual solutions. With a digital computer, however, the more generalized the approach, the better. The results are achieved by following the systematic procedure of suc­cessive substitutions of neutral-axis parameters to ab­solute convergence, which for this example was reached in six cycles (see Note in Example 1). Computer print­out results, including the geometric properties at each cycle, are shown in Figure 6, and diagram of the dis­tribution of the soil bearing under the footing in Fig­ure 7.

LITERATURE CITED •Czerniak, E., "Concrete Support Analysis by Computer," HYDJ<OCAitBOI<

Paoc&SSJNC AND i'ETROLP.UM RF.YJNER, 42 No. 8, 117 (1963). •Czerniak. E .. "Analytical Approach to Biaxtal Eccentricity." Journal of

the Structural Oiv., Proceedings of ohe Amercan Society of Civil Engineers. ST4 ( 1962), ST3 (1963).

Concrete Support Analysis by Computer

Axial loading plus two-directional bending in reinforced concrete sup­ports is an easy problem for a small c~mputer using this simplified program

Ell Czemlak, The Fluor Corporation, Ltd., Los Angeles

HERE'S A GENERALIZED TECHNIQUE together with all the formulas especially developed for the systematic solu­tion by a digital computer of biaxial eccentricity problems in reinforced concrete. The approach is unique because in spite of the length and complexity of the equations, the complete analysis program can be easily crammed into the comparatively little memory space of the small com­puter with a core storage capacity of only 4,000 alpha­merical characters. The program is completely general and can be used for sections with symmetrical as well as non-symmetric steel arrangements, multiple layers of steel, sections reinforced with more than one bar size, unusual modular ratios, rectangular base plates with or without anchor bolts, and to find the maximum pressure under an eccentrically loaded footing with uplift at one comer.

Many constructional components of structures used in the hydrocarbon-processing industry for supporting heat exchangers, accumulators, drums, compressors, piping, etc., are subjected to various combinations of axial loads and bending muments. Because precise analysis, except in the very simple cases, was found to be rather difficult, structural designers in the past had rationalized them­selves into some remarkable oversimplified assumptions that very conveniently bypassed the otherwise tedious solution. Such attitude of "ignore it and maybe it will go away" is both wasteful and dangerous. As a rule, func­tional and more economical, slender structures, built of higher-strength materials, are now used in refineries to support much heavier and larger processing equipment than the massive wall supports of days past. Single col­umn tee-supports and rigid frames, such as seen under construction in Figure 1, when subjected to lateral loading (e.g., from wind, earthquake, impact or vibration) in addition to the equipment weights, often involve the load­moment configurations requiring a stress analysis for axial load combined with two-directional bending.

The Neutral Axis In Reinforced Concrete. The major problem, in both manual designs and in formulating procedures for sequential electronic computation is the determining of the position of the neutral axis, the in-

FIGURE 1-Single column tee-supports and rigid frames.

termediary that is needed before achieving the final re­sults. This computational complexity in reinforced-con­crete stems essentially from the common assumption that part of the section is considered ineffective for design purposes (cracked-section design). Thus, even when the shape of the cross-section of the reinforced-concrete mem­ber might be a simple rectangle, the shape of the con­crete's effective portion (used in analysis) need not neces­sarily be one. Depending on the relative values of applied bending moments to concentric loads, the shape of the concrete section to be included in the analysis could very well vary from a triangle or trapezoid to a full rectangle. The fact that the effectiveness of the reinforcing steel is not always considered constant tends further to compli­cate the analysis.

Stress in Concrete. In reinforced concrete design, the concrete itself is generally not relied upon to withstand much tensile stress. (The reinforced-concrete as a whole though is quite capable of resisting significant amounts of eccentric tension loads as will be shown in Example 2.) It is usually assumed that the tension stresses in the flexural computation are taken by the reinforcing steel, whereas the compression is primarily resisted by the con­crete. According to Section 1109 (b) of the ACI Code*

* Building Code Requirement.l for Reinforud Concrete (ACI S18-56}

77

CONCRETE SUPPORT ANAlYSIS

f

s

y •

OR• c OO•b OS • f 0

FIGURE 2:-ln rectangular teetions, locate origin of coordi· nates in one corner.

some tension stress in the concrete is permitted when, in addition to bending stresses, there also exists direct com­pression and the ratios of eccentricity to depth (eft) is not greater than o/3 in either direction.

Assuming a straight line stress distribution the stress at any point (x, y) in the concrete may be written:

f, =I [1-~- _!.._] x, y o a b

where f0 represents the intensity of stress at the chosen point of origin, and the constants a, b designate the inter­cepts of the neutral-axis line on the x- and y-axP.S re­spectively.

In cracked section designs, where the tensile strength of the concrete is completely neglected, the stresses in the concrete must be assumed to exist only in the compres­sion region. The part of the section, over which f•x, 1

would be negative is said to have thus become ineffective for purposes of analysis. I t is apparent from the stress equation that the region over which f•x, y is negative ex-

tends to all points for which the value : + ~ is larger

than one. It is evident, therefore, that in cracked sections the intercepts a and b can be also used to denote the boundaty line of the concrete's effective section. Conver­gence of the two lines until they almost coincide consti· tutes, for all practical purposes, the solution of the prob­lem. For analytical purposes, the steel can be considered as having been replaced by an appropriate amount of concrete. T he area of this transformed concrete is as­sumed to be concentrated at a point which coincides with the center of the replaced bar. The amount of concrete resulting from the exchange depends on the relative effectivenes attributed to the materials. In the strictly elastic analysis, the modular ratio n is the index to meas­uring the relative effectiveness of the steel over that of concrete. The area of the concrete substituted for each bar equals n times A1• Of course, it presupposes that the bond between all tension and compression bars and con­crete remains intact at all times, and they deform to­gether under stress. In reality, this is not exactly true. There is experimental evidence that the bars in the com­pression region are stressed more than would be indicated by purely elastic considerations. Building codes, allowing for this phenomena long ago, permitted an increase in the stress of the compressive reinforcement. The allow-

78

y

~--------------------~c

• • R 0

I~ a j FIGURE 3-The area under compression is a triangle.

able stress values are well above that which might have resulted from a strictly elastic analysis. Section 706 (b) of the ACI Building 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, and the modular ratio n." How­ever, the use of the 2n is not unrestricted. The code states that compressive stress in the reinforcing should be equal to, or less than, the allowable steel stress in tension. De­noting the allowable tensile unit stress in reinforcement by f 1 the equations governing the stresses in the reinforc­ing steel can be written as:

tensile f • = nf0 [ 1 - : - ~ J compressive £'8 = 2nf0 [ 1 - : - ~ J :::;; f 1

The reader should note that in the case of the com­pressive reinforcement, the bar which is under compres­sion is evidently located in the portion of the concrete which has already been considered effective in the analysis. Therefore, the area of the bar must be sub­tracted from the effective concrete area before computing the necessary section properties. Since this might prove rather awkward, an appropriate correction is made in the transformed area of the steel bar instead. As a com­pensation, the force in the compression bar is reduced by the amount which would have existed (in its place) in the concrete. The reduction equals to the concrete stress times the area of the bar, which is:

f0 [ L --i-- ~ J A,

With the transformed area concept, the correction is accomplished by reducing the effectiveness index m by one. T he area of concrete which is substituted for a bar in compression would be equal to [ 2n - 1 ] or less, times A1. Obviously, the or less applies to those bars whose stress has already reached the limiting tensile stress value. In transforming the tension bars into equivalent concrete, no such reduction applies, since by assumption, they would be located outside the effective portion. However, in the limited cases when tension in the concrete is per­mitted, these bars also displace some effective concrete. Hence. they too must have their areas subtracted or the

I n m

100°/o

Compress eon

v FIGURE 4-Variation of five shape& from triangle to rectangle.

modular ratio modified by using ( n - 1) instead of n.

Capacity of Loaded Section. The magnitude of the largest load which can be sustained at a given location ( witJ:tin the prescribed stress or strain limits) constitutes the measure for the capacity of the section. For any known or assumed position of the neutral-axis it can be determined with ease from the equation as follows:

Eccentric Load P = f [A- Q..y - Qox] 0 a b

Where A denotes the over-all effective area of the cross­section and Qo,., Qor are the first moments of this area about the x- andy-axis, respectively.

In most practical problems, however, the position of the neutral-axis is neither known nor can it be reasonably assumed. Given data usually include the magnitude and the position of the imposed load, as well as the material specifications. The problem then becomes one of deter­mining the adequacy of the section to sustain a given de­sign load, acting at a given point, and not exceeding a given stress limitation. The location of the neutral-axis may be, in itself, of very little interest to practicing engi­neers. Nevertheless, it must be determined first, before proceeding with the more essential task of establishing structural adequacy. The general equation* for the pa­rameters of the neutral-axis are:

x-axis intercept a=

(lxy-Yp~;y) (l.y-XpQo,.)- (lox-Yp~x) (Ioy-XpQoy) (~x- YpA) (I,.1 - XPQ0 ,.) - (Q01 - XPA) (101 - YPQ0 ,.)

y·axis intercept b =

(11 y- Yp~y) (I,.,.- XP~")- (10,.- YPQ0 ,.} (l0y-Xp~y) (Q07 - X 11A) (Ixy- YPQ07)- (Q.,,.- YpA) {I0y- X,~,.)

Where lox and loy are the. moments of inertia about the x, y-axes, and I,.,. denotes the product of inertia of the area about the origin. XP and Y P are the coordinates of the applied eccentric load.

In the above equations, all the section properties obvi­ously pertain to the over-all effective section. The proper­tics of the effective portion of the concrete are added with the transformed properties of the steel.

Rectangular Sections. In the case of rectangular sec­tions, it is convenient to locate the origin of the coordi­nate system in one of the corners of the rectangle (see Figure 2) and let the axes coincide with two sides. The main advantage is the relative ease with which the vari­ous formulas for the required section properties can be

*For background and dcvdopment of th-.e and the other equations listed in this article •cc the author's "Analytical App<oach to Biaxial Eccentric­

ity" Paper 3239 in the Journal of the Sttuetural Divialon, Proceedinso of tho American Society of Civil Engineen, Vol. 88. No. ST4, Augwt 1962.

obtained. Furthermore, by choosing (as the origin) that corner at which the concrete compressive stress is a maxi­mum, the number of possible shapes of effective concrete area is reduced to five.

In Figure 3 the corners of the given section are 0, B, C, and D. Line QR designates the neutral-axis, and in­tersects the X· and y-axi.s at a and b, respectively. When tension in the concrete is not permitted, the neutral-axis line is also taken to represent the boundary line of the portion of the concrete section considered effective in the analysis. When the neutral-axis intercepts are smaller than the corresponding dimensions of the section (as shown in Figure 3) the area of concrete under compres­sion is a triangle.

As one or both of the intercepts are increased beyond the section's dimensions, the effective area progresses from that of a trapezoid to one of a rectangle. When the neutral-axis falls completely outside the section, the whole area is obviously under compression and therefore fully effective. The variation of the five shapes, from tri­angle to rectangle, are shown shaded in Figure 4.

y

-

t d

a

o~d b~t

J X

The required properties of the effective portion of the concrete for the five possible shapes can be obtained from the formulas for shape IV. The geometric properties in terms of the neutral-axis intercepts and the section di­mensions are shown formulated below:

I [ ("-d): (b-t), ] AR.EA="f ab 1- -a- - -b-

1 [ (a-d)' (b-t)' (b-t)' t] Q..=s•"' •- -a- - -b- -s I> 1>

1 [ (a-d)' (b-•)' (a·-d)' d] Q.,=r;a'b 1- -.,.- - -b- --3 -3

- -;

79

CONCRETE SUPPORT ANALYSIS •••

enter with Old poromlfM A and I and section sides C and D

compute .R • T A I

le•t_ w ether \ v.t _, A I . ? I it potttive

J, c •

' no ) rocked'-: ec:tlon

zero out~. e~ate to one:

AR A,~Ooy lox, lov XY

move nro to MARK

move parameter A to AXIS • L

move side D

toWIOTH • W

o.dd one to MARK

test whlrfher \ L- W

Is nevative ? j

no

compute

~.e f-L

~ y ·•oo•t. . compresSion #

move 20 to parameter A

move 2T to parameter 1

subtract ICR from 00y

subtract 2CII(2• Cl

from loy

move parameter I

to L

move side T to w

Y"-i .. ,

~-K•O R•O

compute compute

(J..-i_W)' • I( r.- (!-~w_)'· "

subtract K from AREA

5ubtroet

R from ~ox C1'f

1 subtroetfox

K1

from l:~ J

subtract

4Cit from In

~~·' K for NE

no

test MARK for

TWO

no

~---- ---/ EXIT t~ \ , error routtne 1

.... ----- _.;

aubtroct .. 3Cft from OoK

sublroct 2CR (2t C) from loy

~ compute

Y·G l

multiply AIIU by 0

Ol)d store •n AREA

mult•ply Oox by Gil!

and store in OoK

mulliply QOY b~ GA/3

a d store in Ooy

fS • mumr>•Y 1 lox by G& t&

and store in lox

~ multiply a

Joy by GA i'l .a"\ store •n v

multiply 1 1 XY by (;/12

oM store In 1 xv

EXIT to steel

properties routin •

FIGURE 5-Computer sequence for properties of effective • concrete section.

To compute the contribution of the steel is compara­tively easy. The transformed properties of all the indi­vidual bars are added. Care must be exercised to assign the proper effectiveness index to each bar. In order to differentiate it from the modular ration, let the effective­ness index be designated by m. For tension bars, the numerical value of m is made equal to n in cracked sec­tions and to n- 1, when concrete tension is permitted. For compressive reinforcement, its value is 2n- 1 when the bar stress is less than ft. When the stress in the com­pressive bar reaches (the allowable steel tensile stress) the value of m is reduced to

ft - 1 C::: 2n- l

f (~-~-:!__) 0 a b

where fo is the concrete stress at the origin.

Therefore, the required transformed properties for steel reinforcement are:

Area

MoJllent Areas

80

N

A= r. mA1

i == 1

N

Qox. = r. m Al Y l i == 1

N

Qoy = r. m Ai xi i == 1

N

Moments of Inertia 10x. = .!: m A1 y 12

i == l

N Product of Inertia lx.y = .!: m A1 x1 y 1

i == 1

N

loy ·== 2: m AI xi2

i=l

Convergence Technique. In the usual design problem_, it is necessary to find the size of the reinforced-concrete section which can adequately sustain a given system of loading. Several loading combinations must frequently be considered, and the trial sections must be incremented until all conditions are satisfied. The most laborious part of the computations (as design engineers well know) is determining the parameters of the neutral-axis line. The coordinates of the applied load are calculated from the bending moments (usually given with respect to the cen­terlines of the section). Together with the properties of the assumed section, they are used to determine the neutral-axis parameters. There will be only one neutral­axis which will satisfy equilibrium conditions and stress­strain limitations. When the concrete is not permitted to take any calculated tension, the neutral-axis line is as­sumed to be the boundary line of the effective portion of the concrete. The problem, then, is to find that neutral­axis which almost coincides with the edge of the effective section. The cliff erence betwen the two lines constitutes the measure of the computational error, which, obviously, should be kept as small as practicable.

The work of finding the required parameters may fre­quently be facilitated by following a systematic procedure of successive substitutions until the desired results are achieved. To begin with, the distances of the load from the coordinate axes are determined. With them and the properties for 100 percent compression (with neutral-axis parameters equal to twice the section dimensions) the first trial line is determined. If the neutral-axis line falls within the section, it is subsequently used to define the effective section, all the properties are recalculated, and

About the Author Eli Czerniak is a principal design engineer with The

Fluor Corp., Los Angeles. He coordinates computer applications for the Design Engi­neering Dept., reviews manual tech­niques and develops new methods and procedures better adaptable to

• • • systems conversiOn 1n automating the design and drafting of refinery units. Mr. Czerniak received a B.S. in engineering from Columbia Uni­versity in 1949 and an M.S. in Civil Engineering from Columbia in 1950. He is a registered engineer in Cali­fornia and has published a number of technical articles. He has had field experience as a civil engineer

Czerniak and worked in design and drafting with Arthur G. McKee Co. in Union,

N. J., for two years before joining Fluor in 1953 as a structural designer. He soon headed up the structural design and drafting on various projects until assuming his present position.

new parameters are determined. The process of substitut­ing the calculated parameters of the neutral-axis for the parameters of the effective section is repeated to any de­sired degree of approximation. The convergence routine is quite fast, and only a small number of cycles will usually be sufficient for most practical problems.

Computer Program. When setting up a computer pro­gram for solution of e1')gineering problems, heavy emphasis should be placed on the simplicity of the input data and clarity of the output. With the formulas and procedures described before, the writer developed a program for solving biaxial eccentricity problems, on the basis of elas­tic action, with a small computer, having a core storage capacity of 4,000 alphamerical characters. Because of the widespread availability of these small units it should in­terest engineers that even without Fortran capability they can be used for numerous analytical applications. The program was written in SPS (Symbolic Programing Sys­tem) and punched into 529 cards, which were later con­densed into a 104 card deck. The card reader has a rated speed of 800 cards per minute, which means that it takes approximately 8 second to load the whole program. Com­putations and printout average one-half second per cycle. Absolute convergence, wherein the neutral-axis param­eters {measured to three significant figures to the right of the decimal point) remain the same through two con­secutive cycles is usually achieved within eight cydes. In most instances the results of the third iteration seemed to have sufficed for all practical purposes. In the two ex­amples cited, six cycles were required for the absolute solution. From start to finish, loading of the program and data cards, computations and printout of results for both examples, took 15 seconds.

Now, after the input data has been entered and ma­chine digested, the convergence routine starts with neutral-axis parameters equal to twice the section di­mensions and computes the necessary 'transformed' sec· tion properties from which, together with the load coor­dinates new parameters are calculated and subsequently used. How formulas for the section properties for only case IV are used in the program to determine all possible effective-section properties is illustrated in the block dia­gram, shown in Figure 5.

At each iteration cycle the equilibrium load compati­ble with the section properties, load coordinates, limiting stresses and the newly determined neutral-axis para­meters, is computed and printed together with maximum

SlOE 0 SlOE T lOAD COORDINATES NO. Of URS )( '(

u.ooo 14.000 - 9.000 - 7.000 06

AREA 0 0 I OK OY ox

y

z"tlr r 6-11 .s bars • 1 B'i Co<>r!oles

lTypl 2.375" 2 375"

·-t-~-· 2.375" 11.625" .. f

9.ooo'' 2.375" <IJ

9.000" 11.62!1" • • 15.625" 2.375"

1 e. ·' .I 15.625" II .62.5'' X

I. 18" 0

FIGURE 6--Comer colwnn in exchanger structure used iD examples.

compressive stresses in concrete (at corner used ·for origin} and steel, as well as the maximum stress in the tensile reinforcement.

Numerical Examples. The reinforced-concrete section shown in Figure 6 is a corner column in an exchanger structure. The allowable unit stresses are: 1,350 psi in concrete (for f/ = 3,000) and 20,000 for the reinforcing .steel.

Example I. The column section shown in Figure 6 is loaded with a compressive force of 15 kips, and with bending moments about the centerlines of the section equal to 22.5 ft. kips and 17.5 ft. kips, in the x and y directions, respectively. Determine whether the section can adequately sustain the above loading using effective modular ratios of n equals 10 for the tensile and 2n- l or 19 for compressive reinforcement.

y

• • • -~ .

-(\J . 0

• • X 0 12"

w

-..... J g'

A WEIGHT SUM 0 PIOOULAR RATIOS s II N " 2.640 9.012 l't.Ut. 10.0 19.0

I I PARAMETERS UY XY A II

STEEl 50.1&00 )51. uoo lt!il.lt400 .l510. 79)6 S530.6624 3160.0798 36.000 28.000 CNCR1' 252.0000 1764.0000 2268.0000 16464.0000 27216.0000 15876.0000 TOTAL 302.1600 2115.1200 2719.4400 l99<i4.1936 3271t6.6624 19016.0798 CYCLE CNCRT FO 1350.000PSl STEH,COMPR l9S48 TENSION l60l LOAD 323201 19.200 l5.486NEW

STEEL 34.3200 203.6100 282.6450 1902.9860 :U53.91l7 1770.1817 12.000 10.200 CNCRT 61.2000 208.0800 244.8000 1061.2080 U68.8000 624.2400 TOTAL 95.5200 411.6900 S27.41t50 2964.1940 4722.77l7 2l9ft.4217 CYCLE 6 CNCRT FO l350.000PSI STEEl,tOMPR 15411 TENSION 19454 LOAD 1Sl2U 12.000 10.200NfW

FIGURE 7-Computer solutioa to problem Example 1.

81

CONCRETE SUPPORT ANALYSIS •• . is an actual computer printout. For these eccentricities the load coordinates listed with respect to one corner as origin are: Solution. With P = 15 kips, the eccentricities are:

22.5 X 12 . e. = = 18 mch~~ .• 15

I 7.5 X 12 14

• h e1 =

15 = me es

xp = - 9 inches

YP = - 7 inches

The solution of this prqblem is shown in Figure 7, which

K y 14.64

After convergence, the load capacity is shown as 15,128 lbs. which is slightly more than the given 15 kips and hence O.K . The capacity of the section to sustain this eccentric load was evidently limited by the 1,350 psi compression in the concrete. Maximum tensile stress was 19,454 psi which is close to the limiting 20,000 value given; maximum compressive stress in the steel is 15,417 psi. The final neutral-axis parameten carne out to 12 inches for A and 10.2 inches for B, which means the shape of the effective portion of the concrete section was a triangle (case I in Figure 4) .

( tension) ~------------------~

::

SI DE 0

u .ooo

•

SJDE T

H.OOO

AR EA

30.3600 2 7.8109 58.1109

• •

•

J LOAD COORD INATES

)( y l8, 000 H, 000

Q

OX Q

OY

X

NO . OF BARS

Ob

I OX

Example 2. What is the maximum tension load which can be supported by the section of Figure 6, if the eccen· tricities of Example 1 are halved?

Solution. P = ? With eccentricities equal to e,. = 9 inches and e7 = 7 inches it is apparent that the load is located at the corner of the section, and since it is tension, must be opposite the corner used as origin.

A s

2.640

WE IGHT SUM 0

' I

OY

9,012 llooll6

I XY

~OOULAR RATIOS N 11

10. 0 19 . 0

PARMIETERS A a

7,844 7,0ql STEfl CHCRT TOTAL CYCLE 6 CNCRT FO

l9olj . 2050 65 . 7356

2 S9.9ft06 760, 4'56PS I

24 7,0050 72.1162

319 .7 212 STEElrCOMPR

1880.6491 23J,ObS9

ZlLJ.11SO 5536 TtNSION

2933. 2ll7 28':i . t 930

J218,4047 20000 LOAD

1685.5367 128.9077

18 14.'141,4 - l 4636ll 7.091 NHI

• s . ~40

.<40

·"'0 ·""0 ·••c ·'"0

TOT.t.~~

A s

.. ~~tO -~•o ,.ltttO . 440 .4~0 ..... 0

lOTUS

82

FIGURE 8-- C~mputer solution to Example 2.

TU~SfORM~D PRUP .. ~I lfS OF STEfl Kf !NFn~tJ NG 8A~S (for txomple I}

UUIOEIIA A P&AJoot TE• 8 ..-:lO<Jl AA MU 101 11 ·~n .. HUtl&f R Rf ftAAS U.OD-0 10.200 oo.o ~~.o o .

MOO, AIUA Q I I U< UY ux DY

IS, U~ llo6lS 10, 0 ...... ooo ~ l.J\00 &8 . !SOP Sq~.b\*1 1014.11~1 2. ,, 2.)1~ }9.0 1.!600 J9.n~o I Q. &1SO 4111. 1 'lioS' •1.1 \~{. 2, US 11.615 10.0 4,4000 ''·' soo IO • .C.\0(,1 S"'to\.,U/ H.&lll ,.ooo 2.HS JQ,O •• u.oo & •. 8SSO !~.1.00 u .• , s. 6 1 l. lb0b ·9.000 11.625 10.0 • •• ooo 11. n oo l~ ... ooo Sv~. 6lAl ;so , 4000 u .• zs 1.ns 1o. o 4.4000 Jn.••oo ·~· 1~00 ''·"'~' IOI•. 21R7

WHGHI. SUI< 0 9, 0111 14.Jl6 H . l?Oll 20l.-.oo ?Rl. ~·~n lfJI)l, ')M~O .~)'d.? 1\1

TA.t.I!SFDU£0 P~OOf RTi t 5 01 S I HL Mf 11<fO• C I"~ ~ARS (for exompfe 2)

PA-·HETlK .t. PARAOI£!(k ;; 1400UL A~ KA 1105 N AKU " NU-~f ~ flf 0AA~ 1 ..... !.091 10.0 19.0 0~

• ..oo. uu 0 0 I I u• OY or DY

IS.6H U . US 1o.o • • • 000 S I. ISOO 68. 1500 ~'J4 . 6187 10h . H 8 7 2. ns 2 . llS 19,0 •• )600 l9.HSO 19. ~~50 H.ISS6 4l.J ISL lo )lS 11 , 0 2 1 10.0 ,. • • ooo Sl.l500 10.4500 5?4.6UJ 14. &181 9.000 z.ns 1o. o 4."-000 10 . 4100 39.6000 24. e t&r 156,4000 9.000 11·625 10.0 ~-•ooo Sl .ISOO l9.~U00 S'J4. 4 1ff1 316··~00

IS.6H 2-375 10. 0 •• ~000 IO. •soo b~.HOO ?4.8111 I014.21Kl

~UC11T SUM 0 •• 0121 ~"--11& 10.1600 19<.7050 147.0050 18~0 . 6<~1 z•H·"' '

FIGURR 9-Properties of reinforcing bars for Rxamples l and 2.

XY l99 .. JHil

41.l'> !(l IZ J ,4•11 lli.M~O "-bO. l)OU 161. 2~\l

ll/0. 1~ 11

I l1

199.1117 41.1116

111.461 1 ... osoo

••·O· HOO J6),ZRI Z

Jt~ A 'l.~1nl

The computed tension load capacity is 14,636 lbs., limited by tensile rein· forcing stress of 20,000 psi. (Figure 8) Maximum comer stress in the con­crete (at th1· origin ) is 760.456 psi and maximum stress in compressive reinforcement is 5,536 psi, which as expected is way below the allowables.

The main purpose of the second example is to illustrate the often over~ looked fact that the reinforced con­crete section, as a whole, is quite cap­able to resist significant amounts of eccentric tension loads.

Finally, to show how the individual bars contribute to the transformed properties of the section, a short sup­plementary routine was written that prints out all pertinent properties of the reinforcement for the converged parameters. The computer printed properties of the steel reinforcing bars for both Examples l and 2 are shown in Figure 9. # #

NOTES

FOUNDATIONS t< SOIL~

' . . \. ,

.. . . . .. .

... .

Foundations on Weak Soils

Because today's plants are being constructed on filled sites not ideal for founda­

tions, a careful check must be made on settling tolerance and soil preparation

John Makaretz, The Badger Co., Inc. Boston, Mass.

ToDAv's PETROCHEMICAL plants are being constructed in locations and under conditions that require more at· tention to foundation design than was customary in the past. New plants are often close to water, on filled sites, where the land is not ideal for foundations. In recent years the trend has been to higher towers, often com­bined in groups; equipment has become heavier. More­over riaid reinforced concrete structures permit only

) 0 d . d negligible differential settlement. Tank foun atlons e-

serve particular attention.

New Design Techniques. These considerations suggest the desirability of design innovations or nonconventional design techniques. Foundations having negligible settle­ment can be designed, of course, but their cost is usually prohibitive. If soil conditions permit uniform settlement of two to three inches, however, it is often possible to design foundations at a considerable saving, without sacrificing safety.

It is important to keep in mind that it is easier to predict the settlement of fills, placed over uniform de· posits of clay, than it is to predict deflections of pile foundations loaded by a structure and subject to down­drag load from subsiding fills. An error in the prediction of footing settlement in dense sand is not serious; an error in predicting the behavior of piles in silt clays can result in very serious damage indeed. Foundations on sandy soil will settle quickly and will be stabilized, pro­vided no considerable change in subsoil water level occurs. Foundations on clay settle slowly and over a longer period of time, the settlement also depending upon water level variation, but not to such an extent a.~ it does in the case of sandy soils.

Storage tank foundations appear to be unimportant structures in petrochemical plants. However, considering the large investment in tanks, substantial economy can be realized if, by proper foundation design, long main­tenance-free tank service life is achieved. In order to effect substantial savings on tank foundations, the design engineer and the owner must reach an understanding on

both the tolerable magnitude of settling and the time available for foundation preparation.

Nearly every large tank which is supported on soil will have, after years of service, about one or two feet of differential settlement between the shell and the tank center. The reason for this is the unit soil pressure at the tank bottom. For a tank about 150 feet in diameter and 50 feet high there will be approximately 130 psi under the shell and 23 psi in the middle of the tank.

A large differential settlement between the shell and bottom may cause a tearing or shearing effect l;letween the bottom plate and the shell. However, large tanks over 150 feet in diameter can be used if differential settlement

85

is as large as 24 inches, because of the flexibili ty of the bottom and the roof plates. The effect of the relative settlement between the tank and the connecting pipes can be overcome by using flexible joints. Differentia l

It is easier to predict the settlement of fills over clay than loaded pile deflections subjed to downdrag from subsiding fills

settlement for small tanks (up to about 30 feet in diameter) should not exceed about 1f2 inch. If the dif­ferential settlements under the shell itself are closely spaced, excessive stresses in the shell will occur and the shell may buckle.

Edge Treatment. If the tank site is underlain by a firm subsoil stratum, the following three . foundation methods can be used after the topsoil and organic material are removed:

• Recompact the subgrade and put a pad of sand or gravel directly on the subgrade.

• Use a sand cushion as above with edge protection consisting of a crushed rock ring wall.

• Use a reinforced concrete ring wall, which supports the tank edge, with a sand cushion of about 4 inches inside the ring.

The necessity of using the edge treatment is a con­troversial subject. Some owners feel that "edge cutting" is not detrime1~tal and that the cost of the edge treat­ment is, therefore, prohibitive. Others are of the opinion

86

that the concrete rings are desirable even for the best soil conditions. As arguments for this reasoning, the following points are used:

• A surface level to within ~ inch around the perim­eter is necessary for proper tank erection.

• Even small localized deflection of the foundation during operation may cause "hang-up" of the float­ing roof.

• Edge cutting under the tank shell may cause rupture of the weld between the tank bottom and the tank shell.

• Ring foundations prolong tank life because the edge of the shell is a few inches above exterior grade; cor­rosion problems and maintenance costs will be mini­mized.

• Some tanks need anchorage (aluminum tanks or tall tanks having small diameters).

About the Author

John Makaretz is the chief structural engineer with The Badger Co., Inc., Boston, Mass. He has had a wide experience in structu ral design in building dams, bridges and heavy industrial construction . After receiv­ing an M .S. degree in engineering from Lwow Institute of T echnology in Poland, he practiced structural engineering in Europe for several years. Before joining Ra dger, he was c hief s truc tural engin ee r with Thomas Worcester Co. in Roston. Mr. Makaretz is a member of the American Society of Civil Engineers, the International Associa ti on for Bridge and Structural Engineering Makaretz and the American Con crete Insti-tute. He is a registered _professional engineer in the State of New York, New Jersey and several other statE's.

Weak, Compressible Soil. If the area on which the storage tanks are to be constructed is underlain by weak and compressible soil strata, not over approximately 20 feet thick, the following methods of foundation design can be used:

• If the thickness of weak deposits is relatively shallow ( 3 fo 5 feet), it is often advisable to remove the weak materials and replace them with well-com­pacted granular fills. Note that it is necessary to extend the compacted fill beyond the tank perimeter.

• For deeper, weak soil deposits, it is entirely prac­tical to surcharge the compressible strata before the tank foundation is constructed, if time permits. The purpose of such a surcharge is to increase the st.rength of the subsoil and to reduce the tank settle­ment during operation.

• The tank foundation may be put on a crust of very strong fill and allowed to float on weak soil strata. This is practical where the ground has to be filled anyway. The crust must be thick enough and extend far enough beyond the tank perimeter to prevent lateral plastic flow of the weak subsoils. Steel sheet pilings, concrete rings, or crushed stone rings may be applied to prevent lateral flow of the weak sub­soils which might cause tank foundation failure.

Sheet steel pilings, concrete rings, or crushed stone rings may be used to prevent latera/flow of weak subsoils

Stability analysis of cohesive soils may be made using either A. W. Skempton's method1 or the balancing moments method between the imposed load and the shearing stress resistance of the soil strata in question.

In order to calculate the stability of the tank founda­tion, properties of a clay stratum are required, such as: undrained shear value (C) lbs.fsq. ft., density (y) lbs.jcu. ft., the bearing capacity factor (nondimensional), and the height of the surcharge (d) ft. According to Skempton, ultimate bearing capacity of clays is ex­pressed as:

quit= C X Nc + Y X d (assuming that the clay is saturated, an angle of shearing resistance >f = 0)

The cohesion of the soil (C) for our purpose may be assumed equal to 50 percent of the nonconfined com­pressive strength of the soil. The bearing capacity factor N. varies from 5.2 for elongated footings to 6.2 for round and square footings.

For rectangular footings,

B .!\J,. = 0.84 + 0.16 ~X ~c (for square or circular footings)

1.6"4

c - .... 1.4

g ~ :: (.) 1.3

,;-; 1.2 - .. :{!u > ..5 :•.o =~ 0 09 (1)0

~ \ \ \ \

~' I . I '· ...... 0 ~ o.e

- ll) 0.7 SN I I'·

1 - • .,... I ~·-·-(.) ... 06

wO 1 I 1 ·~·-·-·~ I I 1 I I f= ~ 0.5 I I 1 I I

t.O

25 o.4 Of The Total Plant Cost In Million Dollars

FIGURE 2-Soil inv~tigation cost as a percent of founda· tion, structures and buildings cost.

Where B is the width and L is the length of the rectangular foundation, in feet.

Although it is very important to establish the ultimate bearing capacity of clays in which shear failure may occur more frequently than in noncohesive soils, never· theless the settlement probability for the foundations on clay should be considered and its expected magnitude should be checked. This is especially important if a safety factor of 2 or 3 against ultimate failure is projected.

Example. Applying the dimensions given in Figure I to Skempton's formula,

Nc =coefficient 6.2 for round and &quare footings, 5.2 for strip footings ( nondimensional)

N. (adjusted)= 5.2 + ...!!._ = 5.44 50

quit = 5.44 X 800 + 0. 75 X 120 = 4442 lbs. per sq. ft.

Tank load = 65 lbs. per sq. ft.

Liquid load = 1900 lbs. per sq. ft.

Pad weight= 240 lbs. per sq. ft.

Total= 2205lbs. per sq. ft.

4442 Factor of safety:

2205 = 2.

2) Using the balancing moment method:

rz 2 XC XL quit X-

2- = C XL X r =- ~'

5.44 Reduction factor p, = -- = 0.88

6.20

2 X BOO X 12 X 3.14 qull -- T

2 = 5024lbs. per sq. ft.

Reduced quit = 5024 X 0.88 = 4421lbs. per sq. ft. (the result should be the same as in Case l, or 4442 lbs. per sq. ft.)

Settlement for above conditions. Assumptions: Lw,

87

A pile foundation for tanks with a reinforced concrete slab capping is best but the most expensive; or, a 4-loot capping of crushed

stone compacted between the piles transfers tanlc load to piles well

liquid limit of clay = 40 percent; eo, void ratio = 1.2; Cc, compressibility factor = 0,009 (L,.-10) = 0.27.

It is assumed that this strain is constant for the clay stratum and the clay is normally consolidated (i.e. no drying effect on the surface occurred) . The factors as­sumed above are usually obtained from laboratory tests.

In order to achieve better average conditions for the

Soil investigations are a small part of total plant costs yet

some owners object to taking a sufficient number of borings or any at all

settlement calculation the 12-foot clay stratum is divided into two 6-foot layers.

Approximate ae ttlement {in.) , tl. = H X 12

X c. I P+tt.p 1 + eo ogto --p-

For part a, p = 2' X 120 = 240 Jbs. per •q. ft.

3' X 11 5 = 345lbs. per sq. ft.

585 lbs. per sq. ft.

tl.p = 1900 lbt. per sq. ft.+ 65 lbs. per sq. ft.= 1965 lbs. per sq. lt.

p + tl.p = 585 + 1965 = 2550 lbs. per tq. ft.

6' X 12 2550 tl.a = 2.2 X 0.27 X log1 0 ~= 5.7 in.

For pan b, p = 2 X 120 = 240 lbs. per sq. ft.

9 X 115 = 1035 lbs. per sq. ft .

1275lbs. per sq. ft.

p +tt.p = 1275 + 1765 = 3040 lb1. per sq. ft.

(.O.p for the part b, decreased in accordance with Boussines q formula)

6' X 12 3040 . ~b -

2.2

X 0.27 X log10 ~ = 3.4 m.

Total deflection~ 9 in.

It is assumed for the above investigation that the tank

88

foundation pad was in place at least three months before tank erection.

The above solution to tank support includes con­siderable risk in comparison with pile foundation design. However, the necessity of releveling the tanks several times during installation still may save money as com­pared with pile foundation construction. The conven­tional pile foundation for tanks, with reinforced concrete slab capping, is the best, but the most expensive in comparison with the tank cost. Alternatively, a capping of crushed stone, about 4 feet deep, compacted between the piles, may be used. The compacted crushed stone creates arches between the piles and transfers tank load to piles relatively uniformly.

The exact prediction of tank settlement is impossible, except if supported on point-bearing piles, for the fol­lowin.g reasons:

• The stress distribution in thin, weak soil layers under the foundation cannot be accurately determined.

• The magnitude of lateral plastic flow in highly stressed soils is unknown.

• Behavior of the crushed stone cap on the piles is difficult to predict.

• The nature of the deflection of piles in soft soils is unpredictable.

Relative Cost of Soli Investigation. Let us consider an average size petrochemical plant, the total cost of which is about six million dollars. The approximate cost of foundations, structures and buildings would be about 25 percent of the total cost, or $1 !/:1 million. Soil investi­gation for such a plant would require about 10 borings, which with laboratory analysis and a complete report would amount to from $4,000 to $9,000, depending on soil conditions.

Plotting soil investigation cost against 25 percent of the total plant cost (foundation, structures and buildings) we obtain a curve shown in Figure 2.

The cost of soil investigation is small if it is related to foundations, structures and buildings only; in com­parison with the total plant cost, it is almost negligible.

It is hard to understand why some clients strongly object to taking a sufficient number of borings; some object to taking any at all. A comprehensive soil report enables an engineer to design with confidence, repays the cost of the soil investigation, and saves money for the owner.

LITERATURE CITED 1 SkemptoDJ A. W., ''The BeariD~r Capacity of Clap," Buildizla Reseazeh

ConJlreu, 19;>1. • Proceeding• of the American Society of Civil Engineen, The Journal of

Soil Mechanics and Foundationa Div., Part l, Oct. 1961.

Graphs Speed Spread Footing Design

When designing square or octagonal footings, these graphs will cut the calculation time to a minimum. No trial and error sizing is required

F. 8. van Hamme, Chief Structural Engineer, Fluor-Schuytvlot N. V., Haarlem, Holland*

TRIAL•ANI>-ERROR SIZING of spread footings can be sup­planted by a better method. A graph can be used to size footings with a minimum of calculations. Use Figure l, to determine the size of square footings, Figure 2 for octagonal footings.

Stability and soil bearing are the main considerations in designing spread footings. Equipment must be .sup­ported by the foundation so it will not be overturned by maximum forces acting upon it. The load on the soil

*Subsidiary of The Fluor Corp., Lid., Los Angtl<s, Calif.

12

...... I .,.., ·" J j '

below the foundation must not exceed the maximum allowable soil pressure. The most severe stability condi­tions are realized when the vertical load is minimum and the lateral loads (winds or earthquake) are maximum. Severest soil bearing conditions are realized when the vertical and lateral loads are maximum.

Both graphs were scheduled for an allowable soil pres­sure of 1000 lbs. per sq. ft. However, these graphs can be used for any allowable soil pressure if the vertical load (including the weight of the footing and backfill) and the overturning moment about the base of the footing arc divided by the soil pressure coefficient N.

N = allowable soil pressure in lbs. per sq. ft. 1000 lbs. per sq. ft.

.. 1000

= total vertical load in kips

N

M[QQO =overturning moment in ft.-kips N

Square Footings. When the footing can turn on the A-A-a.xis only (Figure 1) (pipe-rack footings, for ex-

0 1/ II'.

....... (/)

0. ~

' .-II..

I II

--~.,..

v v ~ /, 'V)

>·L~>j

v ~!·' k -- ·-·-- f- tl jf1:!---~ / 1/-

v It l/ / IV ~ !- [i

Ia ~ ~ ,.. _j 'f / ,, '/.A ~ ~ ~

_///.4 v It' I/ ~ / ~ 1/) , ~ v C>. I<:: ~ ~ ~

~ 0 0

=~ 5

-1 .....

~

J

IL ..... Y.. /

~

/ >< /. .?'

1::::: ;;.>" ... .............

V1000 {KIPS)-----

1/ / I' ) v / ......

...., ..... / v V' ..... 1<. ~ / v ~- ....

·il ..... D< / > ... .,, / ~'--... v ~ i/

~5' )<l5'

I"' v ./ / ,... 1"-o / I' v ,. A lA

Or " / / v ~ <.~ .::~ '· '~,·

~ / >< ~ '"' r-... ·~,

~~ ,.., L :>< ,;" IL '~ I' ["\

/ < / ..... < 14 ,, I'. I'-...... .<.. ./ ,;" > ,

~. ... _·.::· ' ...... ..... ,;" ;> ~ ~ "" ' ~ ...... I"'

>C :;;... ..... ...... I~ ..... r .. ' ' ...-.:; ~. ......; 1'. ....... r-... r, " "' ............ --~'-E ·- " ...... I...._ ..... - ...... r--.. I' r' - ~ ... r ....

...... r--. :_, :::.. ~ ~ - f--· f-· -·· s ~ ~ -~ ...... ~ '41

' 10 15 20 25

NOTES• J. Dotted 0-lines ore to be used for stability conditions and the solid D·lines

for soil bearing conditions when footing con turn os well on A·A-oxis osB·B 2. When footing can turn on A·A·oxis only the dotted 0-lines ore to be used for

stability as well os soil bearing conditions.

FIGURE 1-Use this graph to design square footings.

89

--------------------------------------------------- ----· -

oO 0 480

1 I I I 2! I I I I_ I ... II

I I I I I I I I I I I I I I I I I 2 0 . 1 125 2..~ ,~.-303.5 I I I I · ~ f.... 3.0 ,.. 7.,.; 4.555 I -1-f-+11 1 r-...1/ 20 '(/ " 6.0

:Lt r-.. 3.5 U> s.L5 v. ~r/17"1J' ,.., 1 I/ r--.. J I !!:: 11f' ~ f7' .£ ~ I

1/ 1~ ~'I 1 A ~ IXliL ~[/t.,; r<: D-7'6' 400 1 II II II I II I L i rf i'. II .O it IO J' '(f~ t'\..7' · t-

il"' - .L II ~ ..t_ ._. J~li':~ ~ IL i.;' I" X 4.5 0 ~ ......... ~ "" " I

' ~ I' ~ L 'L g 1 ~·~ ~ ' ~

7,{)()

?00

-(/)

a.. 100 :X:: ...,: LL

0 0 0

~

71/ II .L "" ' r\ 5.0 ~ O I I ' ""' I~ N I I I 1111 5,.1.51/ II/_ 1;11'./ J r--.11', 11~~ Ill\. VI ) 5.5 0 10 20 30 40

I J I Ll 11 l Ill I/ ' !1 ........ "- ""- ./1 60 r-r- V1Q00(KIPS) II v ~ '1.1 IL~ IL !t..._L '" ~ ~· · ----l/Vl/1 JIIIT T7l II ~I Y I 1\lTlXTI/I'l VI I'J

I1l11 JLllll~ lll Llll I 111\.1 VI 'XL I lXI I L 1\ RlLJt'l.ZL I TI I T'KfT1 'KI:A I Vl I 1\1. I A 1\ VI/I YJ 1~1 I I A~ I y-r zx-VCI\l!V r;

UMYYIAIIIII IINJLLJ)[IIAILI'W~ ]'\ 111/lii/IA JA"Tl7"to...LIII I X: I V I'lL J' I X I 1/NA I 1-..1 I I" 1\.

MLYY UJrll-Nl ~P'Vr R 1 NIUN] N II'l 1 1 'l"'. ~.,, , , " ..

I ~~tm±§ctl'tli'W\~~Itil

~ ~Ei ~ I~ 0 50 100 150 200 250 300 320

V1000 (KIPS) ___ _

FIGURE 2-Use this graph to design octagonal footings.