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.
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 determi ned as follows: STACKS AND TOW ERS are closely related as far as foundation desi~n is concerned in fact, the same principks apply. Tn 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. SelfSupporting Tower There are two main considerations in designing the foundation for a selfsupporting 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.
w
S,= a(2) where a= area of base of foun dation (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 shoul d 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 onglnrlng 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
5. Overturning Load The overturning load S~ is the result of the overturning moment. Under ordinary conditions, the only for ce 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 baromet ric pressure of 30 inches, the formula becomes : p =0.004V'
(5)
It 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 to wer (pounds per square foot) .
Wind Preaeure Pw
:r:
.. .. c
CJ
~
•a::• Foundation Top
Gra~e ~~'V
Foundation Base
F
Db
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:
I
P .. = p. D.H
~
(8)
L =lever arm of wind load (feet) to be calculated as follows: H L= hr+ 2 (9)
f"A
·
d
FIGURE l Foundotion for selfsupporting tower.
6
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 t o 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:
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 (M 1 ) varies from point to point, and the maximum load (S,) can be calculated as follows : Mr
s.=z
(10) where Z =section modulus of the base of the found"ation. (Not e~ Z to be based on dime nsions in feet .)
The value of (Z) can be expressed as follows: I Z= c (ll) where 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 (5 1 ) and (5 2 ) 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 S 2 ). Since the value of S:~. at point E is negative, the minimum soil load (which obviously occurs at point E) is (S1  S 2 ). 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, .. =  .(2a) 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 ( 1a) Therefvre, in order that (501111 ) may always be positive, thereby assuring a stable condition at all
times, (S1 m) must never be less than (S 2 ). In a perfectly balanced system, (S,m) is exactly equal to (S 2 ) , in which case
s.... = s... S.=o
(1d)
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 s uch cases is the minimum which can exist and still maintain a stable system. It should be emphasized that while (S 1m) 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 t he 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 sliderule 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. = 30,000 631 000 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. L = 6
54 • + 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) _ 200,000 ft. pounds b S2 = 803 I s./sq. ft.= maxi248.5 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.
This loading is satisfactory, as the soil will safely support 2000 lbs.jsq. ft. From equation (2a)
s.... =
125,700 lbs. 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 Wt e= eccentricity (feet)
where
(12)
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 (W 1). It has been shown by previous discussion that the following relationships exist: Wt s... =a
(2a)
Mt
s.=z
(10)
I
Z=c combining equations (10) and (11) :M,c
·=.
S
(11)
(13)
rearranging equation (12) Mr= Wte
(12a)
combining equations (12a) and 13) _ Wtec S2  l
(14)
It was shown by equation (1a) that in order to avoid tensile stress at E (which would make the foundation unstable), the maximum value of (S 2 ) is as follows:
OCTAGOJJ
a = o.a2842. c:
0.54ld
I : O.OStid+ (15)
So= S,.,
Z : O.l016ds
thus making the value of (Smln) equal to zero, as shown by equation (1d). Substitutipg the values obtained by equations (2a) and (14) in equation (15)
r : o.&5'14
w.
w~ec
=a(16) The value of I can be expressed as follows: I
I= ar• where r =radius of gyration of base (feet)
HEXAGOH
a = o.se84'
(17)
c : o.s7'7ct I : 0.064•
substituting in equation (16)
w.ec
w.
ar'

z : o.104cts
a
(18)
r
Hence, the maximum value of (e) for stable conditions is
= o.ae44
r•
e.....
=c
(19)
In the case of a circular foundation d
c=y
a:
(20)
Substituting in equation (19) e.... =
I
2r'
d
r: o.. aaM
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 middlequarter of the diameter of a circular base. In the case of the octagon base usually used for tower foundations, the maximum allowable eccentricity becomes
Mt 200,000 foot pounds 125,700 lbs. e = W, =
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
:"'"D'
· z : o.u8cl,
(21)
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 (S 2 ) 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
o.707
ct•
c:..f1 : o.ot9ct•
z : o.oe8cl'
r=+ FIGURE 2 Eltmectta of foundotion bases (Axis AA)
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) or
S= S;+kS.
(58)
s = s. (1 + k)
(59)
Since the term ( ~) occurs m both equations (65) and (66), it follows that
From equation (2) and
w
s.=a
(2)
From equation (10) (10)
therefore Mr= S.Z
(60)
Substituting the value of (S 2 ) from equation (57) M, = S, kZ (61) therefore
s.=w
(62)
From equations (2) and (62)
and
WkZ M, =a
(63)
(64)
From equation (12)
In the case of an octagonal base, a= 0.828 d'
(70) (71)
z = 0.01016
e= Wt
Therefore, for an octagonal base, equation (69) be written:
m~
(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: W
= 174,200 pounds
(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 press ure) which occurs when the dead load is maximum. It is therefore necessary to substitute (W) in place of (Wt) as follows: Mr e= W
(72)
Mt = 200,000 foot pounds
M,
(65)
Equation (64) can be written
10
(68} Substituting t his value of (k) in equation (59) S= St (69)
s....... = s. ( 1 + 8.15e) d
M,
S,=a = kZ
kZ
w =;
k ~ z
KochJ••=d
Mr
M,
(67)
(t+ C:)
Mr s.=z
W
kZ e=a
(66)
Therefore by equation (65) e=
200,000 174,200
= 1.1
s
The maximum soil loading due to the dead load (S 1 ) 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 ) 1155 ( 1 13.5 S = 1155 X 1.693 = 1955 pounds per square foot.
s=
+
This checks the value of 1958 pounds per square foot calculated (by slide rule) in Section 7, thus
............... c 1·· _.· ·...,.......~ :_jI .......· . 1,: C:::..:.... . . . __j t f ·  ·  · f!.'l·&.. .::• _, . w q, ___ _
"
"'v ~~ ,..... · '
r,
p FIGUU le
fiGUU 3d
FIGUU 3c
indicating that either method yields the same result.
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).
10. Soil Loading at Any Point Having calculated the eccentricity, it is a simple matter to determine t he 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 :
11. Stresses in Tower Shell
S=S.+S•
(1)
Since the value of (5 2 ) for points to the right of the axis is negative, the value of (S) will be: S = S, 
s.
(see equation 1a)
( 1b)
Equations (1) and (1b) can be combined as follows: (1c) S=S,±S, From equation (2)
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 press ure, 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 .. (
_ M,c S'  I
(2)
The value of (5 2 ) can be written: S, =
\Vee' 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 (1a), (14) and (17) referred to above were used to determine t he stability and the eccentricity under the poorest condition, which obviously occurs when the dead load is at its minimum. Equation (24) can be used to determine the stress under any dead load, therefore, equation (24) may be based on either (W) or (Wt) depending on the dead load for which the stress is to be determined. It is
(26)
where M, = bending moment about base of tower (foot pounds). also
w
s.=a
~)
(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 footpound unts. Steel stresses, however, are ordinarily given in pounds per square inch. In order to convert the stress from (pounds/sq. ft.) as calculated, to (pounds/ sq. in.) it is necessary to divide (S,) by 144. T he shell is a hollow cylinder, in which case: D
c=y
(28)
and I
'71'
= 64 (D• D\)
where D = outside diameter of tower (feet) D, = inside diameter of tower (feet) when t = thickness of shell (feet) and D D,=2t D,=D 2t
(29)
(30)
(31)
11
Substituting in equation (29)
The maximum tensile stress per foot of circumference to be resisted by anchor bolts is
'1r
I= 64 [D' (D 2t)']
(31)
Substituting the values of I and c in equation (27)
s. =
D
2 ___ ___M,__; '11'
&f[D' ( D2~)' ] 
32M,D
[0' (D 2t)']
[8D't 24D't'+ 32Dt'  16t')
(32)
The values of t 2, t• and t 4 are quite small and the three terms in the denominator containing t hem may be neglected without introducing appreciable error. For practical purposes, therefore, equation (32) may be written as follows: 5 •=
32M,D 8?TD•t
71'D.
w
(38)
The load to be carried by each bolt can be expressed S.. _ 'lTD• .( 4Mt W. ) 
N
4M,
= NDb
 w
S.. =maximum load on each bolt (pounds).
The nut should always be tight, placing some initial tension on the bolt. Of course due allow(32b)
4Mt
(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. =
(34)
12. Foundation Bolta for SelfSupporting Tower The foundation or anchor bolts for a selfsupporting 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 th e 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 : 4M•
'1rD!,.
(35)
where
Do= diameter of bolt circle (feet).
The compressive stress per foot of circumference due to the weight of the tower is
w. 11'Db
12
(39)
where
T he thickness of shell plate required to resist the bending moment only, is therefore t=
(37)
(32a)
This equation may be reduced to: 4Mt Sa= 71'D't
W. ?TDo
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:
32M,D [D' D' + 8ntt 24D't' + 32Dt'l6t') 32M,D

4M,
(36)
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).
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:
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 or
R.= or
p,. (
1f)
h,
(47)
P,.H
R.= 2hI (48) where h, =height from top of foundation to collar (feet}.
15. Foundations for Guyed Towers It was shown by equation (1) that the total soil 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.)
(40)
=
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 (S,) can be determined as follows: SR. r a
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: (42) R, X cos fl R. X cos 8 (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
(SO)
From equation (2) S.=
Sin 8 R. X cot 8
(~)
w a
(2)
Substituting in equation (49) S= W+R. a
(51)
16. Foundation Bolts for Guyed Towers 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 foll ows: 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 produces a negative bending moment at the collar,
_
... '1 _ ._ ........ _.a."' :or t '
T
.__
..
;!_'
""*" .:. b It'lr ..... 1
~
T I
1I tl.::~
 
I ~i= ·t            1 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,)'
(53)
H )' 2h,
(54)
(H 
P,.H ( .M, =  2  1 
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 (32b) 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
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 o r sinking under the last blow, on sound wood. (inches)
~
.1 t4 r~1 ,r,. Hr
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:
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 (S 1 ) 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 (ZZ 1 ). 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, b 1 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
a, b, u, t, a, t, g b, fu, a, b, u, t, a, t, g b, fu, ·
oqrv oqrv O
(Fl... 3e)
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 dept h 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) :
Unit soil loading
=
63,000 lbs. 32,700 lbs. 95,700 lbs. 95,700 lbs. 151 SQ. f t. =
d,=72". d,=2lw. d.=I62" d.=72±21 ±21 = 114" Line (gf) =67.1" (see Table 1). 114 Line (m, w) 57". . 67.1 X 57 _ , _ (b') Lsne (a, b,) =  47.2 . 81
Factor j = .87 (see Table 2). 803 X 57 = = 5651bs./sq. ft. 81 = 565 ± 522 = 1,087lbs./sq. ft.
Load (m, r) Load (qr)
Calculate shear load (V, ) w X 1,087 lbs./sq. ft. " 47·2 X 24 144 SQ. in.
9.95" X 24" X
S.
~~~
238
47.2 X 24 X 2 X 144 9.95 X 238 X 24 X 2 3 X 144 Total (V,)
8,550 lbs.
1,805
=
935 263
= 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.
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 (ZZ 2 ), (Figure 3c) at the edge of the pedestal. This shearing load may be determined as follows: (81) S.= s.t s. The stresses in this case are not distributed over the foundation area, but are concentrated at the edge of the pedestal. Then
S,
= lineal total maximum unit shearing load, foot of pedestal perimeter).
f
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.
(lbs. per
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 : S, =
633 lb I

7
This stress is satisfactory, as 40 lbs.jsq. in. would be allowed (see Table 2).
The unit stress (diagonal tension) resulting from this vertical shear load can be determined as follows:
Concrete base Earth till Total
162 Line {mw) = z  =81"
W,
± W, ± WP (a, S. u) 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 (S6). The value of (S 6 ) 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
s. = Square
s.=
(83)
Mr .832d.'
(83a)
Mr .943d:
(83b)
Circle
Mr
s.=TBW
(83c)
I n 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'! ~al culate 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. = 48,500 lbs. W.= 119.2cu. ft. X 150lbs. = I7,850lbs.
a.= 29.8 sq. ft.
s.w=
2,000 lbs./sq. ft. 2.484 X 8 = 19.87 ft. 3(),000 ± 48,500 17,850 (29.8 X 2,000) S.= 19.87
L. =
+
= 1.850 lbs.llin. foot.
Calculate shearing stress due to overturning moment (S1 ) by equation (83) M. = 200,000 foot pounds (see section 7).
d."= fl=36..
Sa= ::~~ = 6,820 pounds/lin. foot S, = 1,850 ± 6,820 = 8,670 pounds/lin. foot. d.= 21" 8,670 44 dI . f, 12 X 21 :....: 3 . poun s sq. an.
=
16
6
(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: Sl(n,t() = S.
R:r + 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 q 1 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 t he 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 twothirds of the distance from point (a) to point (t). The individual components and their respective lever arms are as follows:
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 crosssection) 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 inchpounds, as the stresses in concrete and steel are usually given as pounds per square inch, whereas in figuring soil loading footpound 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 • P• J'b•
(85)
where · dr = depth of base, measured from top of concrete to centerline of reinforcing steel: (inches) Mb =bending moment in base. (inchpounds) f. = safe working stress, reinforcing steel in tension. (pounds per sq. in.)
A0•
P• = ( 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 reinforcement 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 centertocenter 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 Line (m2 w)
72"
=2 =
Load (m. r,) Load (q, r,) = Load ( v, p) =
36" 803 X 36 = 357 pounds/sq. ft. 81 357 522 = 879 pounds/sq. ft. 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" . · X 879 pounds/sq. ft. X  2 = 184,000 m.lbs 144 sq. m. 18.65 X 45 45 X 2 = 153,000 144 X 879 X 3 29.8 X 45 446 X 45 X 2 = 62,300 144 X l 318.65 X 446 X 45 45 X 2 144X3 X 2 x3Total (Mb)
= 51,900 = 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 _/ f(baloa. .•l
="'
451,200  10.5" 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). 3 Number required ~i J = 5.5. Use 6 bars within the width of beam (ab). 29 Spacing = 4.96''. Use 5inch spacing en67 tirely across side (gf). which will require = 13 bars per set. Four sets of bottom reinforcing bars will be required for the octagonal foundation.
·t'
·t'
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.= 'dtf o1
(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 (inchpounds). Example No. 7. Determine top reinforcement to resist uplift in the foundation referred to in example No.1. Weight of concrete 1SO 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.
Moment .~ 29.8 X 45 144 ·X 570 X 2
18.6:: 45
X 570 X 45 ~
=
Z
(M~)
Total
119,000 in.lbs.
= 99,500 ;:::: 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 0inch deformed square bars, at toinch 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.
= :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) u
Example No. 8. Check bond stresses in example No.1. Bottom reinforcement
Bond stress for bottom reinforcement is satisfactory, as 75 pounds is permissible (see Table 2). Top reinforcement TAILI 2 Conttcaftta Applying to foundation DNI8" Mt.ture:
Cement ..............................
Sand .................................
Coane Aallftaate .....................
1 l 5
I l 4
500 376 2,000 1.500 800 600 
fb
Safe bcvlna load on concrete (lbs./IIQ. in.).
f~
Ultimate compreuive etrength (lbt./IIQ. in.)
fo
Safe unit atress In extn:me fibl!t' of concrete (in comprcsaion)~(Jba./sq. ln.) ......•..
f4
Safe unit strass In concrete due to vertical shear (diaronal tenaion} (lbs./~q. in,) ••
f.
Safe unit stresa in concrete baae due to punchina shear. (lb&./IIQ. in.) ••........
120
f.
Safe workinll, atreu. steel reinforcement in tet~tion. (I e./tq. in.) ..................
18.000
(f. j)
(.l.bchpounds) ..........................
16.600
(1. Jp.)
(l11chpounde) ..........................
138.7
I
Ratio. lever arm of rnlatlng couple to
depth (dr) •••.•••.••••..•••.••.••.•..
.87
Efu.. I
Ratio, modulue o( elatticity of steel to that of concrete ..........................
16
[ n ..
{p.•Ao/
I
/b.dr
•u
Ratio. effeetlve area of tenalon reinforcernent to elf ~tive area of concrete ......
40
.0089
Safe bond ttrea.t (concrete to steel rein· forcemeDt} per "Ualt of area of ~urface
of ~:in(f:~~~·! ...•••................ Deformed bart .....................
60 76
30
90 18.000 16.000 88.9 .89
IS .0056 45
56
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
=4,480Jbs.
= 947
Total (V.) ::t. = 3 X .5 X 4 = 6 5427 u = 6 X .87 X 22 =48tbs.
= 5,4271bs. (88)
The bond stress in top reinforcement is satis
   factory, as 75 pounds would be atlowed.
• Theae tiguree may be allihtly lno:r~ by makin& "U"beoda on thf'
ende of tbe relnfordnl ban.
Figure shear 47.2" X 24" X 570 lbs. 144 9.95 X 24 X 570 144
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.
(37b)
(pound s/sq. 1n.)
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, 'iT
fb
n,•
= 
+ R.+w.+w. 71' n,
(37c)
12r.,
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 reinforcedconcrete 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) F or 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 o f slab to resist uplift struses:
M. Ao "" f,jdr a~
8' ) (ta
area of base of foundation (sq. ft.)
= 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
&v
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 t o resist th e diagonal tension stres ses. ( line &1 bs, Figure 3.) (inches) c  distance from neutral axi s of foundation b"
=
19
d,
= short
P .. 
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
•' ' or e ....   c ema.: 
(19)
z
(19a)
a
Values of (emu) for various foundation shapes are as follows: (22) Ocl&Jon,: e ....  O.l22d (22a) Hexagon: e ....  0.121d (22b) Square: e ••,  O. ll8d d (22c) Circle: emaa8 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). barometric prusu.re. (inchu H 1) 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.) unit stress in concrete (in diaronal tension) due to vertocal shur load. (pounds/sq. in.) unit streaa in concrete base due to punching shear. (pounds/ sq. in.) F I 
r.
1
fv 
s.
i2dt
(68)
Valuea of k for variou• foundation shapes a1·e as follows: 8.15e Octaron: k d(72) 8.32e
= (1 k = ..!:!~ d
IIe:ucon : k Square: Circle:
k
=
S.Ole d
(72a) (72b) (72c)
lner arm of wind load (feet) t o be calculated as follows: II
2
L~hr+
(9)
(
~)
(26)
liLa  bending mom.:nt in base, due to uplift forus. (inchpounds) 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
= ( A.of ba dr)
R•
= pull on
JUY wire due to wind pressure. (pounds)
R• 
Re Sin I
(40)
or,
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:
(1 7)
r~! rw
(25)
~
width of tower base rinJ. (Inches) unit soil loadinr. (pounds/sq. ft.) (I) SSs also, S St (I+ lc) (59) s, =unit soil loAdinr due to dead load. (pounds/sq. ft.) ~ 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~ s tal 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)
Vu = (90 f plus 150 D) (B x) = 555 X 3.3 bond stress
= 1,850 lbs./ft.
. 1 850 0.867 X 17 X 1.6 = 79 pslok
Concrete design for 2,500 psi at 28 days f0 = 1,125 f, = 20,000 adding 33y; percent overstress for ACI 603(c) f 0 = 1,500 f, = 26,667 .iFor D = 1.75; let d = 17 inches; C' = 0.91 X 9.25 = 8.42
Example (2)0klahoma. A5811955 gives a 40 PSF pressure zone. Wind pressure 0  30 feet 30 X shape factor 0.6 = 18 PSF on vert. proj.
;For '10 < 3, proceed as follows:
30 50
40 X shape factor 0.6 = 24 PSF
X= 1.5 (B2.) = l.5 X 10.8 = 16.2
50100
50 .X 1hape factor 0.6 = 30 PSF
Y=
66
~
(BC' d/6)
= ~ (19.58.42 2.83)=4.13
100up
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)
Shear stress S,
156 265 • = 99 9 PSI< 100 (101 plus 28) X 0.867 X 14 ·
= 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 = WP = 70.84 X 4.5 X 150 = 47,820 lbs. ..; 82 = S1 = 1,501 PSF
438,420 lbs.
(2 BS 3 BZC' plus C'S) 486,240 lbs.
Weight of base 182 X 1.5 X 150
Wb=
72,900 lbs.
Weight of soil above 90X3.5 (32471)
W8 =
79,700 lbs.
Operating load on soil
w
0
=
638,840 lbs.
W 1 = 28,800 lbs.
deduct
W min= 610,040 lbs.
Minimum direct soil load Water to fill
Ww =
244,500 lbs.
Max. test load on soil
Wt =
854,540 lbs.
Wt/82 =
Soil pressure under test load
610,010 X 18 '~miD= 2 X 1,801,200 638,840 X 18 2 X 1,801,200
2,637 PSF
;;;;:: 652,000 lb/ft < M,.r
Seismic factor 0.025 gives M.
1,801,200 , = 2.82 feet 638 840
Consult curve for.!:. for '10 = 3.19 and obtain.!:.= 0.914 p'
. Sot) pressure =
4W 0 ( _ eo) 2 38 8
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 1,853 1,501 Mm = (18  8.42)2 plus X (11,6648 18 24 8,175 plus 595) = 17,256 plus 17,518 = 34,774lb. ft/ft
p'
• 0
4 X 638,840 p; 54 (185.64) P _
X
12 Mm "{' Sb X .867 d
Steel in bottom =
4M 4 X 34,774. V m+ B _ 10 C' = _ _ :;;;;: 14,490 lbs./ft l:0 ·= 5.4 mches 18 8 41 260
X
12
Bond stress =
X
50,960 > M, i.e. no compression 12 steel
J..!.!L: =
vm
0.867 dl:0
Note: if s ... had been <
14,490 0.867 X 14 X 5.4 = 221 PSI< 267
Ys S1, allowable bond stress =
200 PSI
UpliftSince 7JmtD > 3, there can be no uplift. It is usual to provide nominal No. 4 @ 12 inch centers in such cases in the top of the mat, both ways.
1
= 4,189 PSF _ 0 914
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
Example (3)Central Texas. A5811955 gives a 25 PSF pressure zone. Wind Pressure: 0  30
20 X shape factor 0.6 = 12 PSF
30 50
25 X shape factor 0.6 = 15 PSF 30 X shape factor 0.6 = 18 PSF
50100
40 X shape factor 0.6 = 24 PSF
100up
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 feet6 inch
1,801,200 X 6 = 1,853 PSF (18)a B2 Max. shear V D= c1  K) [3 S1 (K plus 1) plus 2 s... (K 2 12 plus K plus 1)] S = M ... r X 6 w 81
Note: if s ... < Ys S1 , then let S,. = 0 and allowable shear stress S, = 75 PSI if S,.
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.
D= 1.5
f=3.5
W 0,. =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(27271) W,=63,320lbs. Operating load on soil W0 = 610,820 + B2 = S0 = 2,244 PSF Operating liquid loaddeduct  28,800 lbs. Minimum direct soil load WmiD= 582,020 lbs. W., = 244,500 lbs. Water to fill Max. test load on soil
wt =
826,520 ..; B2 =
st =
3,036 PSF
67
Seismic factor 0.025 gives M 8 = 651,960 < M,.t Wmla X B 582.020 X 16.5 '~min= 2 M wt = 2 ·X 1' 109, 580 = 4·33 17 o =
A! '10
>
1.5
610,820 X 16.5 2 X 1·,109,50 = 4·54
For example (1), M" = 0.1 (438,420 X 56.75 2,600,470 lb. ft. WR =
Soil pressure (on comer)
_ (48/104) X2,600,470  483,520_ 716,700 26,667 X 32  26,667 X 32 = 0.84 in 2
1,109,580 X 6 = 1,482 PSF (16.5)8
+
1.414 S,. = S0 = 2,244 2,096 = 4,340 PSF
+
Subtract weight of original soil at given depth 90 ·X 5 Max. soil pressure Concrete design for
2,500 psi
 450 3,890 PSF < 4,000 PSF
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
483,520
A
> 3.6, proceed as follows: M,.t X 6 S,.= BS
cw. + Wp + WL) =
+ 45,100 X 2.25) =
i.e., provide 2 at No. 9 and 2 at No. 8 at each of 8 pedestal faces.
++
L D 56.5 (H/ 2) (H/ 2 L Mwt = 5s Mwt
For example (2), M" =
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. A
+ Wp ) = 457,440 373,890 26,667 X 32 = 0.44in•
(48/104) 1,801,200  457,440 26,667 XN
).e., provide 4 at No. 6 at each pedestal face For example (3), . 2 _ (48/104X 1,to9,580  457,440 _ A26,667 :X 16  0.I 3 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 Then Root Area A =
(48/Db) M'  Wa f• X N
WR is usually the empty weight W•• but when the earthquake moment is used, WR is the operating weight W• WL For example (1}, Using SAE 4140 bolts with allowable stress 30,000 psi
+
About the Author
J. A. A. Cummins is a civil engineer working in design and construction for H\ldson Engineering Corp., Houston. He attended 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 195154 he was ·a concrete structural engineer in England, and from 195456 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
M' = 438,420 X 52.5 = 2,301,700 lb. ft.
A_ 
623,900 . 600,000 = 1.04 m2
(48/104) 2,301 ,700438,420 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.
Use [email protected] 11,4inch dia SAE 4140 For example (3) M' ~ A=
~; X 1,109,580 =
(48/104) 1,013,930409,620 12,000 X 12
1,013,930
58,350 041. 2 144,000 = · 10
Use 12 @ 1 inch dia. carbon steel. ( Ys" dia would do here, but normally less than not be used)
inch would
##
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 loadsvertical, lateral, and overturning momentscan 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 offcenter (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 !Computerdesigned 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 centroidwithout 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 makeup of the critical loadmoment combinations could be of economic significance. Figure 1 shows several computerdesigned 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 longduration vertical loads. It should apply only to such loading
71
CALCULATE FOOTING SOIL BEARING . . .
FIGURE 2Spread 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 momentson 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 contribution can be evaluated from the simple flexure formula
~e , provided that the resultant eccentricity e (computed from ~) falls within the kern of the footing area. By superposition, the maximum and minimum pressures are simply the algebraic additions of the direct . components, A+ P d P Me andbendmg Me  an A    , respec1
1
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 closeup 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 agendathe 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 loadmoment 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.
A 
IX
+
Mye 1
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 (neutralaxis) 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:
:j
D ~
D
4
4
~I..
i
y y
T
1
Q'
T
.a
o~ a
j_l o~a ~
1
x
1
T
~T'
.D
.a
ol a
J
j1v ol a
X
l
l
X
1
a
FIGURE 3Depending 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 
~x
C!or a
where A is the effective footing area; Q 0 • and Q 07 are the first moments of area A about the x and y axes; a and b are the intercepts of the neutralaxis 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 crosshatch regions) are shown mapped in Figure 3. ,
Stability. For the resultant load P to be within the kern, the sum of the eccentricityratios in the x and y directions must be equal to or less than onesixth, .
E,.
t.e., 0
+ TE
1
~
1
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 average 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 eccentricityratios 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
8 COMPUTE GROSS AREA: 1        , A6 = DXT COMPUTE MAX. SOIL BEARING: GIVEN P p0 A _ OoY _ Oox a b
=
= IXY Y,.Ooy k 2 = lxY X.. Oox k = lox Y,.Oox k4 = loy X.. Oov ks = Oox YpA k = Oov X.. A 3
6
COMPUTE p, =Po[ 1
COMPUTE:
k,
~]
PRINTOUT: PROPERTIES, PARAMETERS, % OF GROSS, CYCLE
PRINT HEADINGS AND GIVEN INPUT DATA
COMPUTE LOAD COORDINATES: X,. = D/ 2  Ex Y,. = T/ 2  Ey COMPUTE INITIAL NEUTRALAXIS 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
0
1
PRINT: SOIL BEARING AT FOOTING CORNERS
COMPARE PARAMETER b TO OLD b
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 4Logic pattern for computer program.
74
SOIL BEARING ANALYSIS OF RECTANGULAR FOOTI NG Exa mpl e 1
FIGURE 6Computer 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 Pr ogram. 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 reinforcedconcrete, 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 neutralaxis 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 flowcharted in Figure 4. From start to finish, load ing of the program deck and data cards, computations and the printout 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 heatexchangers is
shown in Figure 5. Determine the maximum soilbearing 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.
= 250kips E = 1'·0" E = 1'0"
P
X
y
t
I
0
2
'
(D
.
.
!Ey
p
1~+
1o, ..
t....
I
Ex
.
I
.....·    15 I  0 II FIGURE 5Example L
75
CALCULATE FOOTING SOIL BEARING
 
,..._.,t.~
1"1•¢ ..;;;.;
 
footing dimensions, see Figure 4) , the correct location o{ the neutralaxis 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 neutralaxis, 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.
= 100 kips ~Ex= 1.00 + 2.75 = 3.75 ft. 1Ey = 1.00 + 2.25 = 3.25 ft.
FI GURE ? 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 neutralaxis 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 neutralaxis 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
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: xdirection 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 neutralaxis 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
Concrete Support Analysis by Computer
Axial loading plus twodirectional 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 nonsymmetric 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 hydrocarbonprocessing 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 higherstrength materials, are now used in refineries to support much heavier and larger processing equipment than the massive wall supports of days past. Single column teesupports 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 twodirectional 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 1Single column teesupports and rigid frames.
termediary that is needed before achieving the final results. This computational complexity in reinforcedconcrete stems essentially from the common assumption that part of the section is considered ineffective for design purposes (crackedsection design). Thus, even when the shape of the crosssection of the reinforcedconcrete 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 reinforcedconcrete 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 S1856}
77
y
CONCRETE SUPPORT ANAlYSIS
~~c
f OR• c
s
OO•b OS • f 0
•
•
I~
• y
0
R
j
a
FIGURE 3The area under compression is a triangle.
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 t he concrete may be written: f,x, y =Io
[1~_!.b._] a
where f0 represents the intensity of stress at the chosen point of origin, and the constants a, b designate the intercepts of the neutralaxis line on the x and yaxP.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 extends 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 t he 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. T he 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 A 1 • Of course, it presupposes that t he 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
able stress values are well above that which might have resulted from a strictly elastic analysis. Section 706 (b) of the AC I 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 straightline 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•
compressive £' 8
= nf =
0 [
2nf0
[
1
:

1
:

J ~ J: ;
~
f1
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
100°/o Compress eon
m
n
I
v
FIGURE 4Variation 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 neutralaxis it can be determined with ease from the equation as follows: Eccentric Load P = f 0 [A Q..y 
a
Qox] b
Where A denotes the overall effective area of the crosssection and Qo,., Qor are the first moments of this area about the x andyaxis, respectively. In most practical problems, however, the position of the neutralaxis 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 neutralaxis 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 neutralaxis are: xaxis intercept a= (lxyYp~;y) (l.yXpQo,.) (loxYp~x) (IoyXpQoy) (~x YpA) (I,.1 
XPQ0 ,.)

(Q01 
XPA) (101 
YPQ0 ,.)
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 neutralaxis, and intersects the X· and yaxi.s at a and b, respectively. When tension in the concrete is not permitted, the neutralaxis line is also taken to represent the boundary line of the portion of the concrete section considered effective in the analysis. When the neutralaxis 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 neutralaxis 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 o~d b~t
X
y·axis intercept b =
Where lox and loy are the. moments of inertia about the x, yaxes, and I,.,. denotes the product of inertia of the area about the origin. XP and YP are the coordinates of the applied eccentric load. In the above equations, all the section properties obviously pertain to the overall 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
d
t
(11 y  Yp~y) (I,.,. XP~") (10 , .  YPQ0 ,.} (l0yXp~y) (Q07  X 11 A) (Ixy YPQ07 )  (Q.,,. YpA) {I0 y X,~,.)
J
a
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 neutralaxis intercepts and the section dimensions are shown formulated below:
I [ ("d):  a   (bt), ] 1 [• (ad)' (bt)'s (bt)' t] Q..=s•"' I > 1> 1 [1 (ad)' (a·d)' d] Q.,=r;a'b .,.  (b•)'
AR.EA="f ab
1
b
a 
b
 b  3

3

;
*For background and dcvdopment of th.e and the other equations listed in this article •cc the author's "Analytical App
79
N
N
CONCRETE SUPPORT ANALYSIS •••
Moments of Inertia 10 x.
= .!:
m A 1 y 12
loy ·==
Old poromlfM A and I and section sides C and D
.
subtract K from AREA
compute .R • T A I
5ubtroet
le•t_ w ether \
v.t _,
A I . ? it potttive
y ·•oo•t. . compresSion
I
J,
' no )
Y·G
l:~
1
J
out~. e~ate to one:
zero
lov
move
to
l
multiply AIIU by 0 Ol)d store •n AREA
subtract 4Cit from In
move 2T to parameter 1
AR A,~Ooy
XY
MARK
K for
NE
mulliply GA/3 a d store in Ooy
no
subtract 2CII(2• Cl from loy
move parameter A to AXIS • L
mult•ply Oox by Gil! and store in OoK
~~·'
subtract ICR from 0 0 y
nro
QOY b~
move parameter I to L
move side T to w
test whlrfher \
Y"i
L W
Is nevative ? j
fS
mumr>•Y 1 lox by G& t& and
store in lox
no ~
o.dd one to MARK
•
test MARK for TWO
move side D toWIOTH • W
loy
~
K from

/ EXIT t~ \ , error routtne 1 .... _.;

..,
~
multiply a Joy by GA i'l store .a"\ •n v multiply 1 1XY by (;/12 oM store In 1xv
~
K•O R•O
EXIT to steel
no
properties routin compute
compute
~.e L
f
(J..i_W)'•
I(
compute
•
r. (!~w_)'· "
FIGURE 5Computer 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
f0
(~~:!__) a b

1
C::: 2n l
where fo is the concrete stress at the origin. Therefore, the required transformed properties for steel reinforcement are:
r. = r. N
Area
A=
mA 1
i == 1
MoJllent Areas
Qox.
m Al Yl
i == 1
80
r. N
N
Qoy =
i == 1
= .!:
m A1 x1 y 1
i == 1
compute
subtroetfox
•ec:tlon lox,
from
1
#
Product of Inertia lx.y
aubtroct 3Cft from OoK
2CR (2t C)
C1'f
move 20 to parameter A
crocked':
N
sublroct
~ox
R from
~
xi2
i=l
i == l enter with
2: m AI
m Ai xi
Convergence Technique. In the usual design problem_, it is necessary to find the size of the reinforcedconcrete 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 neutralaxis 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 neutralaxis 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 neutralaxis 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 clifference 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 neutralaxis parameters equal to twice the section dimensions) the first trial line is determined. If the neutralaxis 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 neutralaxis 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 onehalf second per cycle. Absolute convergence, wherein the neutralaxis 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 neutralaxis 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 effectivesection 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 neutralaxis parameters, is computed and printed together with maximum SlOE 0
FIGURE 6Comer 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 reinforcedconcrete 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
•
•
0
X
12"
w
J
.....
g'
A
s
2.640
WEIGHT
SUM 0
II
9.012
I I 0 OK OY ox UY S530.6624 50.1&00 )51. uoo lt!il.lt400 .l510. 79)6 252.0000 1764.0000 16464.0000 27216.0000 2268.0000 302.1600 2115.1200 2719.4400 l99
0
is an actual computer printout. For these eccentricities the load coordinates listed with respect to one corner as origin are: xp =  9 inches
CONCRETE SUPPORT ANALYSIS •• .
Solution. With P = 15 kips, the eccentricities are: 22.5 X 12
e. = .• e1
15
=
18
.
mch~~
YP =  7 inches
• h = I 7.515X 12 = 14 me es
After convergen ce, 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 t he 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 neutralaxis parameten carne out to 12 inches for A and 10.2 inches for B, which means the shape of the effective portion of t he concrete section was a triangle (case I in Figure 4) .
The solution of this prqblem is shown in Figure 7, which
y
14.64
K
( te nsion) ~~
•
•
•
::
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?
•
X
=
J SI DE 0
SJD E T
LOAD COORD I NATES
u .ooo
H.OOO
l8, 000
)(
AR EA
H, 000
Q
30.3600 2 7.8109 58.1 109
STEfl
OY
TOTAL
CNCRT FO
CYCLE 6
760, 4'56PS I
WE IGHT
s
SUM 0 llooll6
I
I
OY
OX
~OOULAR
XY
1880.6491
2933. 2ll7 1685.5367 23J,ObS9 28':i . t 930 128.9077 J218,4047 18 14.'141,4 ZlLJ.11SO 5536 TtNSION 20000 LOAD  l 4636ll
24 7,0050 72.1162 319 .7 212 ST EElrCOMPR
RATIOS
N
'9,012
2.640
I
Q
l9olj . 2050 65 . 7356 2 S9.9ft06
A
Ob
OX
CHCRT
NO . OF BARS
y
=
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.
10. 0
11
19 . 0
PARMIETE RS A a
7,844
7,0ql 7.091 NHI
FIGURE 8 C~mputer solution to Example 2. TU~SfORM ~ D
P&AJoot TE• 8
UUIOEIIA A U.OD0
•s . ~40
.<40
·"'0
·""0 ·••c
·'"0 TOT.t.~~
MOO,
SUI< 0 14.Jl6
1.!600 4,4000 4.4000
11. n oo Jn.••oo
H . l?Oll
20l..oo
P~ O O f RTi t
s
.. ~~tO ~•o
,.ltttO . 440 .4~0
.....0
• IS.6H U . US 2. ns 2 . llS lo )lS 11 , 0 2 1 9.000 z. ns 9.000 11·625 IS.6H 2375 ~U C11T
lOTUS
•• 0121
..oo.
uu
1o.o 19,0 10.0 1o. o 10.0 10. 0
• • • 000 •• )600
0
u•
•• ~000
10.1600
19<.7050
4."000 ~•ooo
I
ux
DY
IO • .C.\0(,1 !~.1.00 l~ ...ooo
Sv~. 6lAl
1014.11~1 •1.1 \~{.
H.&lll
u .• , s.
·~· 1~00
''·"'~'
?Rl. ~ · ~n
lfJI)l,
~ARS
KA 1105 N 10.0
')M~O
IZ J ,4•11 lli. M~O
;so , 4000 IOI•. 21R7
"bO. l)OU
.~)'d.? 1\1
161. 2~\l
ll/0. 1~
11
(for exompfe 2)
"
AKU 19.0
I
NU~f ~
flf
0AA ~ 0~
I
I
~ 'J4 . 6187
10h . H 8 7
H.ISS6 5?4. 6UJ 24. e t&r
4l.J ISL 14. &181 156,4000
or
DY
l9. ~U00
S'J4. 4 1ff1
316··~ 00
b~.HOO
?4.8111
I014.21Kl
147. 0 050
18~ 0 . 6<~1
z•H·"' '
FIGURR 9Properties of reinforcing bars for Rxamples l and 2.
82
XY
l99 .. JHil 41.l'> !(l
6 1 l. lb0b
0 OY 68. 1500 19. ~ ~50 10.4 500 39.6000
SUM 0 ~"11&
HUtl&f R Rf ft AAS o.
4111. 1 'lioS' S"'to\.,U/
140 0UL A~
S I. ISOO l9.HSO Sl.l500 10 . 4100 Sl .ISOO IO. •soo
,. • • ooo
~~.o
Sq~.b\*1
&8 . !SOP I Q. &1SO
5 01 S I HL Mf 11
!.091
1 .....
..
·~ n
UY
''·' soo &•. 8SSO
•• u.oo
• •• ooo
(for txomple I}
I
U< ~ l.J\00 J9.n~o
PARAOI£!(k ;;
PA·HETlK .t.
A
oo. o
Q
AIUA
......ooo
TA.t.I!SFDU£0
8 A~S
..:lO
10.200
IS, U~ llo6lS 10, 0 2. , , 2.)1~ }9.0 2, US 11.615 10.0 ,.ooo 2.H S JQ,O ·9.000 11.625 10.0 u .• zs 1.ns 1o. o
WHGHI. 9, 0111
PRUP .. ~I lfS OF STEfl Kf !N Fn~ tJ NG
l1
199.1117 41.11 16 111.461 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 a nd 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 foundations, 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 ) . denegligible differential settlement. Tank foundatlons 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 maintenancefree 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 flexibility 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 "hangup" 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 fro m Lwow I nstitute of T echnology in Poland, he practiced structural engineering in Europe for several years. Before joining Ra dger, he was c hief s tru c tural en g in ee r with Thomas Worcester Co. in Roston. Mr. Makaretz is a member of the American Society of Civil Engineers, the International A ssocia ti o n for Makaretz Bridge and Structural Engineering and the American Con crete I nstitute. 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 wellcompacted 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
~
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.... 1.4
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 ..
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=~
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0.7
+
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, .!\J,.
Cost In Million Dollars FIGURE 2Soil 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)
quit
quit=
I'·
I
0.5
N. (adjusted)= 5.2
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:
' ·......
1I I
wO ~
.
I
SN (.) ... 06
f=
~'
+ ...!!._ = 50
5.44
= 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. Factor of safety:
4442
2205
= 2.
2) Using the balancing moment method: rz
quit
X2
= C XL X r =
Reduction factor p, =
qull 
2 XC XL ~'
5.44 6.20
  = 0.88
2 X BOO X 12 X 3.14 T2
= 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 4loot capping of crushed stone compacted between the piles transfers tanlc load to piles well
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
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 pointbearing 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.
settlement calculation the 12foot clay stratum is divided into two 6foot layers. Approximate ae ttlement {in.) , tl.
X
c.
1+
eo
I P+tt.p ogto  p
= 2' X 120 = 240 Jbs. per •q. ft.
For part a, p
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.
tl.a
6' X 12
=
. 22
X 0.27 X log1 0
2550
~=
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) ~b 
6' X 12 . 22
• The nature of the deflection of piles in soft soils is unpredictable.
= H X 12
3040 X 0.27 X log10 ~
.
= 3.4 m.
Total deflection~ 9 in.
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
ConJlreu, 19;>1.
It is assumed for the above investigation that the tank
88
BeariD~r
Capacity of Clap," Buildizla Reseazeh
• 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 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.
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, FluorSchuytvlot 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.
N
1000 lbs. per sq. ft.
..
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
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*Subsidiary of The Fluor Corp., Lid., Los Angtl
12
= allowable soil pressure in lbs. per sq. ft.
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NOTES• J. Dotted 0lines ore to be used for stability conditions and the solid D·lines for soil bearing conditions when footing con turn os well on A·Aoxis osB·B 2. When footing can turn on A·A·oxis only the dotted 0lines ore to be used for stability as well os soil bearing conditions.
FIGURE 1Use this graph to design square footings.
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(KIPS)_ _ __ FIGURE 2Use this graph to design octagonal footings.