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FOUNDATION DEffi HANDBOOK =-'
Reprinted from
HyDRocARBo[Fnocisslrvc
rye,?;.7ry.'
;$^w* W
.
Gutf pubtishing
D*.,-tl,
FOUNDA
Published b This reference manual has been reprinted from the regUlar issues of HYDROCARBON PROCESS+ruO. Otfrer- Handbooks Manuals in the series are:
LINES FOR BETTER MANAGEMENT I
N MANUAL
NSTRUM $tr' :l:i,
TME
D.ELARE STA
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;
FOUNDATION DESIGN HANDBOOK TABLE OF CONTENTS TOWER FOUNDATIONS .. Foundation Design For Stacks And Towers Simplified Design For Tower Foundations . Calculation Form For Foundation Design Use Graph To Size Tower Footings Simplified Design Method For lntricate Concrete Column Loading Unusual Foundation Design For Tall Towers Foundation Sizing Simplified . . .. Dowel Sizing For Tower Foundations . . Short Cuts To Tower Foundation Design VESSEL FOUNDATIONS . Foundation Design For 8-Legged Vessels Pressure Vessel Foundation Design . . .
.
,
.
.
COMPUTER FOUNDATION DESIGN How To Calculate Footing Soil Bearing By Computer
Concrete Support Analysis By Computer . .
FOUNDATIONS ON WEAK SOILS
....
.
.
Foundations On Weak Soils Graphs Speed Spread Footing Design Use Graph To Analyze Pile Supports .
.
Page No. 4 5 2',|
27 35
39 44 50 53 57 60
.61
.63 .70 . 7',1 .77 .84 .85 89 .93
\
3
TOWER FOUNDATIONS
#**.*"
-i
{}-@
Jng* dd-P
Foundation Design For Stacks and Towers The same principles ,pply in both stacks and towers. Use this method in making your calculations for either.
V. O. Morsholl, Tennessee Easlman Company, Kingspori,
Frorn the viewpoint of the f oundation designer,
stacks and tor.r.ers may be divided into tr,vo general classifications, deper-rding on the method utilized to maintain them in a vertical position; (a) Selfsupporting, which resist the overturning forces by the size, shape and weight of the foundation ; (b)
Guyed, in which the c,verturning forces are resisted by guy u,ires. It is obvious that the conditions alTectir-rg the design of founclations fc.,r these tr,r,o types u,ill not be the same, ar-rd that it is necessary to treat them separately. STA(--KS AND TOWERS are closely related as far as founrlation design is conccrnecl-in fact, the same principles apply. In the case of stacks, the
brick lining is a r,ariable load, corresponding to, and requiring the sanrc treatnrent as the liquid, insr-rlation, etc., in a tou.er. This cliscussior-r will be based on the design of tower forrndations, however, it should be kept in n-rind that it is also applicable to stacks.
3. Soil Loading (See Section 20 for complete definition of terms.)
The soil loading may be determined following formula:
S: S,* S, where
S: : Sr: 51
by
the
(l)
total unit soil loading (lbs.,/sq. ft.) unit soil loading due to dead load (lbs,,/sq. ft.) unit soil loading due to overturning mo,ment (lbs.,/sq. ft.)
4. Dead Load The dead load S, may be determined as follows:
^w a
Jrwhere
a: 'W:
-W: Wt :
Q)
area of base of foundation (sq. ft.)
total_
weight on soil (pounhj) calculated by
YY.,:l
W.
the following equation:
(3)
Minimum dead load (pounds), which is ihi
weight of the empty tower plus the weight of the foundation, including the earth fill on top
of the
2. Self-Supporting Tower There are two main considerations in designing the foundation for a self-supporting tower; (a) soii loading (b) stability. The foundation must be of such size and shape that the load on the soil below will not exceed the maximum load which it will safely support. The foundation must also maintain the tower in a vertical position, so that it will not be overturned by the maximum forces acting upon it. No direct method of calculating the size of the foundation has been developed, therefore, it must be determined by trial and error. A size is assumed, and the soil loading and stability calculated. If the results are not satisfactory, another assumption is made, and the calculations repeated.
Tenn.
base.
W.: Weight of auxiliary material and equipmenf supported by the tower and foundation
(pounds), which should include the liquid in the tower, insulation, platforms, pipin!, etc. (Does not include weight of tower)
Thls is o revised orticle whlch wos prevlously published
in the August, 1943 lssue of PETROLEUTI REFINER. All copies of thot lssue, oll reprints ond qll coples of thc l94O Process Hondbook, In which the origlnal qrticle wss reproduced, foiled to meet th. demqnd for thls englneering doto. When the outhor consldered tho revlslon he extended
the subiect to lnclude octuol deslgn of foundotlon typ6r commonly required In the erecllon of refinery vcsgctg.
fhe resuli ls o thorough study of o sublect whlch contlnues ro hold o forefront posiiion in refinery cnglneerlng. Reprinrs will be provided ln quantity sumcient to lnclude lhe demond thot hos extended Into constructlon fietds outslde of refining. Prlce $l.OO por copy.
5. Overturning Load The overturning load S, is the result of the overturning moment. Under ordinary conditions, the only force tending to overturn the tower is the r.vind pressure.
The magnitude of the wind pressure is obviously a fun-ction of the wind veloiity, which varies in different localities. In many instances laws have been enacted which state the wind velocity or wind pressure to be used for design purposes. The United States Wcather Bureau has proposed the following formula : p
:
0.004
B io v'
(4)
where -
wind pressure on a flat surface (pounds,/sq ft') : baromctric pressure (inches Hr) V : velocity of wind (miles per horrr) For a barometric pressure of 30 inches, the formula becomes:
;:
B
P: 0'004V'
(5)
It has been found that the wind pressure on a cylindrical tower is about 60 percent of that on a flat surface. For a cylindrical tower, theref ore, formula (5) becomes: (6) P": 0.0025v' where on the projected area of a cylinP": wind oressure(pounds per square foot). drical'tower
Wind Prcssurc
In most localities, zL wind velocity of 100 miles per hour is considered the ma'ximum. This giv.es 'u pr..rrr." of 25 pounds per square foot on the -the
tower, which is the figure generaliy used for design pu.rposes. I.t should be Emphasized, however, tiiat this figure is subject to variation in d.ifferent localities, and that local laws should not be overlooked in this connection' (Note : As a matter of interest,. the wind pres,ri. o., an octagon shaped stack is considered to be ?0 percent of that on a flat surface.). Figure 1 represents a .tower, mountg9 on a concrete-foundati,on. The 'arind pressure (P*) tends to rotate the tower and foundation about point A at the intersection of the vertical centerline and the base of the foundation. This rotating effect produces an overturning moment which can be calculated as follows: (7) Mr: P- L
oriiected area of
Foundotion ToP
where
M, : overturning moment about the base of the foundation (foot Pounds) p* : ioiat wina Laa (bounds) to be calculated
Gro
follows:
P- : o. D"H
Foundotion Bosc
as
(8)
L: ierrer irm of wind load (feet) to be calculated as follows:
H
(9) L-h,* , D.:diameter of tower measured over insulation ( feet) H -height of tower (feet) hr : height of foundation (feet)
It
FIGURE I Foundotion for self-supporting tover'
6
should be noted that alt dimensions are stated
i.,-ieet, giving the overturning moment (M3) in ioot pou"'at. ft',i. avoids the use of the excessively larse numbers encotlntered with the usual inch poind units. Care should be taken, however, to
use consistent units, that is foot pounds,
calculations.
in
all
load, o1-lhe soil resulting from ..Jl-: :l."rs, ormoment (y,) valies from p"oint
:1:_?":.t,i'ning potnt, and the maximum load (Sr) can lated as follows:
bi
to
calcu_
^Mr >2:z where Z
(
:
10)
section modulus the base of the found.ation. (Note: Z to be -of based a;-er.ion. i, i;;i.i
""
The value of (Z) can be expressed as follows:
,:* (11) where : I moment of inertia of ttre base of the foundation (based dimension, i,i-tJ"ti. c: distance fromonneutral a*ii oJ iJiria"tion l"s"
to point of maximum ,t..ss (fiefj. Having calculated- (Sr) and (Sr) as explained above, the total soil ioid ,na")-iilriimum dead
times,.(Sl.) must never be less than (Sr). In a perfectly bal_anced system, (S._) i; .q"ri to (Sr), in which case "*ritty Snro : SrSz:0 (l_d) a.bala".S..d sy:19* is ..Althou.gh,-such rarely pos_ sible, it is the ideal condition.' tt e- up*a.d'fo.ce at E due to the overturning -o*.r,i il-;_;;i; balanced by_the dead load, io tfrui-tfr. stress at r, ls zero. I he stress at F in such cases is the -maintain minimum which can exist ,iiif a
""a stable system. be emphasized.tha_t while (S,*) is fre_ ^__l!:,!.rld qu_enily greater than (Sr) it should nevei be less.
It should also be eri-rphasized tfrai stabilitv is based on the minimum dead ioaa ttre iwJ-;;tia the.soil -loading is based on the ma*)mum dead load (W).
.
load conditions can
b.e
determin.d
bt
This maximum soil load occurs 'r,t"q;"ii;"*(-ii if,. .agd ,;f designated as p, i. irequEntty ll:lgl,"1",ion, ,.toe reterred to as the ""Jft is pressure.,, obviout that the maximum toe pressure (S) should never exceed the safe bearing load of th; ;il in question.
I,
It
should
t. ,,o,"t,ilitigi is positive at point
,ng negative at E (Figu.l'f y.'tn oihe, *brds, rne wtnd load causes compressive stresses on the A, ihe *u*rn"* compres_ :gil t. the left of point and tensile stresses of equal 9:.Yt.ing.at f, :191 magnitude to_the right of A, the maximum tension
occurrrng at II.
Since the earth has no strength whatever in it is obvious that the suir of the stresses at any -point must be positive. fn other words, the base of the foundation. must exert a-comp.es.i.,re force on the soil over its entire ur"i, ott d.*ir" n tensile stress will be produced at E, which *"u.,i that the tower and fbundation wili be unstabte, and .likely to be overturned by the uiiion of the tension,
wind.
.Lt-*": shown by equation (1) that the maximum soll load. is equal to (S, + Sr). Since the value of 5, at .pornt.-b is negative, the minimum soil load (*jr:l obvlorsly occurs at point E) is (Sr_Sr). rs very lmportant to note that the condition ot-rtpoorest stability occurs immediately after the tower is mounted on the foundation, and before the insulation, platforms, piping, UquiJ, etc., are m ptace. In calculating the stability, 'therefore, (S,) must be replaced b"y (S,-) a, foiio*,,
Sr-: Wt a where Sr-: T_r.inimum-
(2-a)
-
S,
__ .421.2 cu. ft.
\
150 lbs.,/cu. ft.
Volume of earth'fill
if-='f
W:i{fu vv erg_nr or
:
tower
+
:
63,000
30,000 lbs.
+
3Z,7OO:125,700 tbs.
Insulation, platforms, piping,
(
l-a)
Therefore, in order that (S*i,) may always be positive, thereby assuring a'sti6fe .onaiiio., ut
"tt
16..
LIJ'rt'r:,1,.!'i;r'.',3,2,11?=39lr36io'l
_e-mpty
30,000
:63,666
W, will be as follows:
The minimum soil loading which can ever exist, therefore, is found to occuf at point E when the dead load is at its minimum value, and can be expressed as follows: Sr-
Since the ,r.., ,,,1""11tti'r,. o",o* grade; the foundation will be 6 ft. deep, with the- top 1 ft. above grade, making the bottom of the foundation 5 ft. below 'grade, br 1 ft. below the frost line. The foundation will be octagon shape, which is recommended for such cases, as it combines the features of stability, ease of construction and mini_ mum material better.than other shapes. The top course will have a short diameter of O ft. sin.l the tower is 4 ft. dia. and allowance must be made for founda.tion bolts, etc. The short diameter of the Dase wlll be assumed to be 13.5 ft. The thickness of the base.will depend on the bending and shear_ l"C l.or.9: (see Sections .1-9 to 19h incl.), however, for the time being the thickness will be assumed to be 2 ft. Th_e weight of the foundation will be calculated as tollows (all slide-rule figures): octagon : 0,8f8 d,^1-0.828 X 6,:29.8 sq. ft. *l:l :f 6.ft. vofumeot top course : 4 ft. X 29.9- 119.2 cu. ft. r\rea ot base (octagor). (q): q.-g?g I3.5,: l5l sq. ft. Volume of base :2"1t. k i5i :30r;;lX i;. " Total. volume :119.2 *' jOZ: 4Zl i ;;.' i;. welght of concrete:
_wt
soil loading due to dead load (lbs.,/sq. ft.)
S*r,:
7. Example No. I Design the concrete founation for a tower 4 ft. dia. by 54 ft. high, including a 4 ft. skirt, and weighing.30,000 lbs. empty. The insulation, platforms and piping weigh 9b00 lbs., the maxirirum wind velociJy is 100 mlles per hour, and the frost line at the location of the proposed'installation is 4 ft. below grade. The maiimum safe soil loading is 2000 pounds per square foot.
etc. :
Water required to fili it t6i"".(4 ft. dia.) (50 ft. high) "
9,000 lbs.
:39,5001bs.
Total (W.) :4g3mlE; W:125,700 f 48,500: LZ4,2OO lbs. (from equation 3)
a: l5l sq. ft. 174,200|bs. ^ -1Sl:qlTI5': :
1155 lbs.,/sq.
ft.:
Maximum dead
load on soil (equation 2)
Allowing 2,, for the thickness of the insulation,
7
the effective diameter of the tower exposed to the action of the wind is 4.5 ft. A wind velocity of 100 miles per hour is equal to 25 pounds per sq. ft. of projected area. Therefore:
p.:25 lbs.,/sq. ft. 4.5 ft. H:- 54 ft. P*: 25 X 4.5 X 54 : 6080 lbs. (equation 8) hr: 6 ft. 54 ft. (equation 9) L:6* 7:33 : 6080 X 33:200,000 foot pounds (equation Mr Z: 0.1016 d' : 0.1016 X 13.53 :248.5
maxi-
mum soil load due to overturning moment'
The total maximum soil load (toe pressure) can
be calculated from equation (1) as follows:
:
1155
* 803:
1958 lbs.,/sq.
TABLE Elements Short Dlam. (Feet)
3.5 4
4.5 5
5.5 D
6.5 7
l.c
Area a
(sq. Ft.)
2. 691
46. 5
3. 105
40.5
52.8 59.5
9.5
74.5 82.8
I
10
10.5
66. 8
sl.2
12
100.0 109.8 119.5
12.5
140. 0
11
11.5 IJ
2.484
34. 8
8.5
8
1.863 2.O70
2.898
3.3r2 3.519
3.726 3.933
4.140 4.347 4.554 4.761 4. 968
r29.2
5.382 5.775
13.5
151.0
14 14. 5 15
162. 0
5. 796
174
6.003
186 199
0. 210
16 16. 5 17
2t2
226 240
77.5 18
18. 5 19 19. 5 21
21.5
23.5 24
24.5 25
25.5 26
26.5 27
27.5 28
28.5
3.78 4.06 4.59 4.87 5.13
74. tO 87. 30
3.5i
5.41,
101.60 117.60
5.94 6.22 6.48 7.03 6.76 7.30
154.10 774.50 227.00 198.00 248.50
5.b/
8.11 8.38
6.831
8..65 9.O2
342. 375. 416. 455.
6.624 7. 038
283 299
7. 866
8.073 8.280 8.487
8.694 8. 901
9.108 9.315 9.522 9.729 9. 936
497
10.143
518 539 560 582 603 526 650
10. 350
10.557 10. 764
10.971 11.178 11.385 17.592 11.799
9. 19
9.46 9.72
10. 55
731.00
to.25 10.81 11.08 11.35 11.62
11. 90
14. 01
14.88
15_ 15
15.42 15.68
17.3t
36 38 39 40
958 1015 1075 1134 1195 1260 7325
16.77 17.85
18, 38 18. 92
15.732
19.47 20.01 20.55
16. 146 16. 560
21 .65
14.904
15. 318
1085. 00
1400.00 1490.00 i585. 00
13.25 13.52 73.79 74.o7 14 .33
8.244
34
933. 00
1005.00
12. 98
72.43
30
13. 662 14. 076 14. 490
873.00
1145. 00 1240. 00 1320. 00
t2.420
902
811. 00
12. \7
15. 96 16. 23
12.434
497.(n
543.00
10.00
12. 006
1,2.2t3
00 00 00 00
590. 00 624. 00 652. 00
720 745 755 848
135. 00
277 .OO
695
31
52.00 62.70
6.417
.245 7.452 7.659
672
16.45
21.90 27.90 34.90 42.80
29
29.5
s.23 12. 68
309.20
7
348 365 383 401 420 438 458 177
.
7. 83
268
20
20.5
5.589
2.43 2.70 2.98 3.25
21. 09
Radius of o.772 0.900 'r
1.89 2.16
\3.2 16. 8
Bose
Sectlon of slde I to Extreme (Feet) lFlbffc(Feet) Modulus Z
7.45
l5r
1685. 00 1787, 00
1900.00
020
1.158 1. 286 1. 415
t.542 1.67
Usually it is found that the first assumption as to foundition size is not correct, in which case, another assumption is made, and the calculations repeated.
it is interesting to note that the soil loadinq
4000 lbs./sq. ft. Care should always- be taken, t.o ascertain the actual load carrying value of the soil at the site of construction.
8. EccentricitY
3021.00
2.44 2.57 2.70 2.83 2.96 3.09 3.47 3.60 3. 86
3.99
4.24 4.37 4.50 4.63 4.76 4. 89
5.02 5.14 5.27 5.40 5. 53 5. OO
5.78 5.92 6.04 6.17 6. 29 6. 43 6. 56 6. 68 6
.82
7.46 7.58
7.71,
3330. 00 3660. 00 3980. 00
a,23
4730. 00
9.26
4370.00
5130.00 5580.00
6020. 00
6500.00
vertical load. As explained previously, it is not necessary to calculate the eccentricity. in order to determine the stability- of the foundation. Several methods have been proposed, however, which make use of the eccentricity, and since there are terest.
2.t7
7.20
2740.OO
be noted that there are two forces acting on foundations of the type under consideration; (a) The dead load, acting in a vertical- direction; (b) the wind load, actingln a hotizontal direction. it e combined ac[io., oithese two forces, that is, their resultant, has thq same effect as an eccentric
2.3r
.93
6. 94
2600.00
of
ft. allowed in this problem is rather low, as good clay soil will usually su-pport abou't 2000 lbs./sq.
2.06
1
2110.00 2220.00 2470.OO
tbs./sq. ft.
definite relationships between eccentricity and stability, they will be explained as a matter of in-
1. 80
2010. 00 2350. 00
:830
stable.
It will
LenAth lNeutral Arls
10.6
20.7 25.0 29.8
ft.
I
of Octogonol
125;700 lbs.
points, thus indicating that the foundation is
7)
From equation (10)
S
Sr-:
This is the dead load under the worst stability condition, and since it is greater than the overturning stress (Sr:803), the soil below the foundation-will always be under compression at all
D.
,,: l@,9%*j$@ :803 lbs./sq. rt. :
This loading is satisfactory, as the soil will
safely support 2000 lbs./sq. ft. From equation (2-a)
8. 48
a.75
9. 00 9. 51
L77
10. 04 10. 28
The eccentricity can be calculated as follows:
e: Mr where -W; e: eccentricity
(12) (feet)
Note : The value of (e) calculated by equation (12) is the maximum value, as the dead load-.(Wt) is minimum. The eccentricity for other conditions of dead loading may be obtiined by substituting the proper weight in place of (W,). It^ha; been ihown-by previous discussion that the following relationships exist:
^Wt -f5r-: ^Mr }z: _7_
z- cl-
(2-a) ( 10)
(11)
combining equations (10) and (11) cJ2-
M'c ( 13)
I
rearranging equation (12) Mr: Wte combining equations (12a) and 13) cJ'--
(12a)
Wtec I
(14)
It w.a_s shown by equation (1-a) that in to avoid tensile stress at E (which would
order
ocTAOOII
make
a = o.gzgdlc : o.s4td
the foundation unstable), the maximum value of (Sr) is as follows:
r = o.o55d+ z I o.Ioledr r: 0.257d
Sz: Sr* (ls) thus making the value of (S-sn) equal to zero, as shown by equation (1-d). .-Substituting the values obtained by equations (2-a) and (1a) in equation (15) Wtec Wt la ( 16) The value of I can be expressed as follows: rT
^-2 al
r:
radius of gyration of base (feet)
where -
uE)ucou
d=
(17)
substituting in equation (16) Wtec W, ar' a H_gnce, the maximum value conditions is
c = 0.577d
rI
(e)
for
o.o6d'
z = o,1o4d'
(18)
of
O,866d,2
r:
stable
o.e64d
r"
enax: I
(
le)
SAUAE
In the case of a circular foundation C:Td (20) Substituting in equation (19) 2f '-"" - d (21) the value.of (rr) in equation (21), ., Substituting the maxrmum value of (e) for a circular base is (*), ,n"r confirming the commen rule that in a stable foundation the resultant must fall within the middle-quarter of the diameter of a circular base. In the case of the octagon base usually used for
d= d2
c:
I TI' z : o.Ilgdt r: 0.289d OIRCIA
q: o.?g6{d2 c -.8-
tow.er foundations, the maximum allow;ble eccen_
tricity
r : o.o{gd+ z : o.og8dt 13-?r A
becomes
e^^,:0.122
d
(22)
The area surrounding the center of the base, within which the resullant causes a compressive
stress over the entire base, is known as the kernel or kern. . ,{t f9l.19ws, th.en, that the resultant must always
fall within the kern of the base in order to stability.
Eremenrs
llnll"loi,,*
(Axis A-A)
r**
assure
No. 1 (Section Z), it was shown that . In-exampl.e the foundation is stable, since the overturnins
stress (!r) i. less than the minimum dead loaE stress (S.1"). The_stability of this foundation will now be cal_ culated_ (as example No. 2) on the basis of the eccentricity for the purpose'of comparing the two
methods.
From equation (12) the eccentricity is
Mr e:=w;:
o.?o7d d4
200,000 foot pounds I25,700 Ibs.
-- 1.59 ft. Frorn equation (ZZ), the maximum permissible eccentrlclty 1s e^",: 0.722d: 0.122 X 13.5 : t.S{ 11. Ipasmuch as the actual eccentricity (1.S9) is less
than the maximum p_ermissible eccentricity (1.64) the foundation is stable, thus confirming ihe corr'_ clusion reached in Section ?. 9. Method of Calculating Soil Load From
Eccentricity calculate the soil loading (toe pressure) as a function of the eccentricity: This method will.b_e explained in order that it inay be compared with the method described in Seciions 3, 4 and. 5. Let (k) be a factor by which the dead load press.ure must be multiplied-to equal the soil loading due to overturning as follows I kSr:Sr (s7)
It is
possible
to
9
.)c ----J
FIGURE
FIGURE 3b
I
Substituting in equation (1) S: S,*kS,
or
S:S,(1+k)
(s8) (se)
From equation (2)
w c-.....:_ rr_ a
u,-
(2)
Mr-
Mr:
(10)
SrZ
(60)
Substituting the value of (Sr) from equation (57)
Mt-S,kZ
(61 )
therefore
^MrTZ Sr:
(62)
From equations (2) and (62) and
r'"-w-Mr a tvlr-a -
kZ
wkz
(63) (64)
Equation (64) can be written M, .W-_.kZ a
IO
(67)
ea k:- -Z
(68)
1-
-
In the case of an octagonal a:0.828 Z
:
(12)
(6s)
d'3 d8
(70)
(71)
Substituting these values in equation (68) 8.1 5e
koct.ro": --6-
U2)
Therefore, for an octagonal base, equation (69) may be written:
/ 8.15e \ s,(r+?/
(73)
For comparison, the maximum soil loading in example Nb. 1 will be calculate4 (as example No. 3) on^the basis of eccentricity' From previous calculations, it was found that: 200,000 foot pounds 174,200 pounds
Therefore by equation (65) 200,m0 e:-ffi:1.15
The maximum soil loading due to the dead load (Sr) was found to be 1155 pounds per square foot, and (dr) is equal to 13.5 feet. Substituting in equation (?3)
/ 8.r5 x 1.15 \ S:1155(r+ff/ S
(66)
base,
0.01016
Mt: W:
The value of (e) calculated from equation (12) is maximum for'any particular foundation, which is the value governing s_tability. At the present moment we are concerned with the maximum soil loadins (toe pressure) which occurs---when the dead l"oad is maximum. It is therefore necessary to substitute (W) in place of (W,) as follows: Mr e: w-
e:-;KZ
soctsson:
From equation (12) M, e: wi
in both equations
Substituting this value of (k) in equation (59) / ee (69) S:S,\r+ zl)
Z
therefore
Since the ,.trn (#) occurs (65) and (66), it follows that and
From equation (10) c-
n
:
1155
X 1.693:
1955 pounds
per square foot.
This checks the value of 1958 pounds per square foot calculated (by slide rule) in Section 7, thus
/l
_._.___:t.-.
)c
---]n i
,-lt
FIGURE 3e
indicating that either method yields the
same
result.
10. Soil Loading at Any Point Having calculated the eccentricity, it is a simple matter to determine the unit stress on the soil at any point whose distance from the centroidal axis
is (c'). The unit stress on the soil, from equation (1), is as follows: (l) S: S, * S, Since the value of (Sr) for points to the right of the axis is negative, the value of (S) will be :
(1-b) S: S,-S, (see equation 1-a) Equations (1) and (1-b) can be combined as
obvious for this purpose that the same value of the dead load should be used in the calculation of the eccentricity (e), by means of equation (12). 11. Stresses
The steel shell is required to withstand the (a) the internal pressure; (b) the dead load; (c) the overturning moment due to the wind pressure. This discussion will be confined to the stress resulting from the wind stresses resulting from,
pressure.
It may be assumed, in determining the stress due to the wind pressure, that the tower is a vertical beam, and that the wind produces a bending moment. The ordinary formulas for determining bending moment and stress may therefore be applied, as follows :
*': n- (+)
follows:
S:SrtS,
(
1-c)
Mtc ^ __l_ S,: (2)
^
W a
(see equations 14 and, 17)
(23)
Wec' Zt'
Simplifying:
t:+(,*5)
Mt: bending moment about base of tower (foot
also
The value of (Sr) can be written: ^ -Wec' Sr: ^l Thereiore,
(26)
where
pounds).
From equation (2)
-_w ", -;
in Tower Shell
(24)
This value of (S) is the total unit stress at any point whose distance (in feet) from the centroidal axis is (c'). It is important to note that equations (1-a), (1a) and (1?) referred to above were used to determine the stability and the eccentricity under the poorest condition, which obviously occurs when the dead load is at its minimum. Equation (24) can be used to determine the stress under any dead load, therefore, equation (24) may be based on either (W) or (Ws) depending on the dead load for which the stress is to be determined- It is
(27)
where St
: unit stress in tower shell due to bending moment (Mt). (lbs./sq. ft.) Note: The unit stress in the tower shell (S,) is calculated in pounds per square foot in order to be consistent with the other calculations which are in foot-pound unts. Steel stresses, however, are ordinarily given in pounds per square inch. In order to convert the stress from (pounds/sq. ft.) as calculated, to (pounds/sq. in.) it is necessarv to divide (S,) by 1aa. The shell is a hollow cylinder, in which case:
c:-TD
Gg)
and
l:
64
where D
: Dr: when t:
and
(D'- D't)
outside diameter of tower (feet) inside diameter of tower (feet) thickness of shell (feet)
- Dr:11 Dr:D -2t D
Q9)
(30) (31 )
II
The maximum tensile stress per foot of circum-
Substituting in equation (29)
I: 64 [D'-
(D
(27)
St:
#rr'-
I
and
c in equation
D
32M.D
2
(D-2t)41 32
:
:@
(31)
-zt)'l
Substituting the values of Mt
ference to be resisted by anchor bolts is
32
- ,]D.-(D]ZDT
:
,rDt
-T-
MtD G2)
(38)
The load to be carried by each bolt can expressed
:r: q--
,lrDn
/ 4M,
-N \ 4Mt :ND"-
z.Do,
W. \ - rD, )
W"
(3e)
N
where
Sr: maximum load on each bolt (pounds).
gTpq-
(32-a)
The nut should always be tight, placing some initial tension on the bolt. Of course due allow-
4M.
z'Dl
(32-b)
The thickness of shel1 plate required to resist
the bending moment only, is therefore r
4Mt
(33) - z.D,St By multiplying the stress in pounds per square foot (S,) by the shell thickn.ess (t) the stress per foot of circumference is obtained as follows:
,S,:#
(34)
12. Foundation Bolts for Self-Supporting Tower The foundation or anchor bolts for a self-supporting tower are required to resist the overturning moment (Mr) resulting from the wind pressuie, after allowance has been made for the resistance offered by the weight of the tower. Obviously the t'esistance offered by the tower's weight is least effective when the minimum weight is acting. The anchor bolts should therefore be calculated for the condition existing when the tower is empty and without insulation, platforms, etc. This weight will be designated by (W"). It was shown that the maximum stress per foot of circumference due to the wind moment can be calculated by equation (3a). That equation gives the stress at the circumference of the shell, however, at the present moment it is desired to determine the bolt stress making it necessary to substitute (Ds) in place of (D). The sffess per ioot of bolt circle circumference can th€n be written: 4Mt
,rDF
(3s)
where
Dn: diameter of bolt circle (feet). The compressive stress per foot of circumference due to the weight of the tower is
.
t2
be
32MtD
This equation may be reduced to:
^5t:
rbsented as follows
MtD
The values of t2, t3 and ta are quite small and the three terms in the denominator containing them may be neglected without introducing appreciable error. For practical purposes, therefore, equation (3?) may be written as follows:
^s1:
4\{t W" (37) TDJ - qrDb Assuming that the number of bolts is represented by (N), each bolt will be required to carry the stress over a portion of the circumference rep-
____--1.
--------.,L-A-
l' \
\
Sl\..
--.LL/-ll./z'.r. /t'r'/ /
\ /',/ .,' ,'
Yi{iii i|i$i 4"M N.
E
t
.t
SLOPE
-r-t :ll lll
iH
'Tl
--'T',---___L----
(36)
O DRAIN
r| ccour
CHAMTER
l.lr
Llt
ltl
Llr
iil
x--6:ro'
'W"
;D;
ry,
FIGURE 4
ance for the initial tension should be made in determining the size of the bolt, and the strength of the bolt should be based on the area at the r6ot of the thread. An additional allowance, usually r/s", should be made for corrosion. The number of foundation bolts should never be less than 8, and should preferably be 1p or more, as the larger the number of boltj, the better the stresses are distributed, and the less danger resulting from a loose nut on one bolt. The bolt should be embedded in the concrete foundation in such a manner that the holding power of the concrete will be at least equal to thE full strength of the bolt. It is common practice to use a washer at the lower end, or to bend the end of the bolt to form an "L" for the purpose of anchoring the bolt in the concrete (iee'Figure ?).
fn
18. Guyed Towers cases where the tower is very high,
it
Pull on Guy Wires pull on the guy wire occurs when the wind blows along that wire, and each wire must be designed to take care of the entire wind reaction at the collar. - -Th" pull on the guy wire can be expressed as follows l 14.
_ The. maximum
The value of the reaction at the collar (R") may by calculating the moments about the base of the tower (the-top of the foundation). The wind moment was found by equation (2d) b.e determingd
to
be
(+)
"-
. The resisting moment arm at the collar is hr, therefore the reaction (R") may be calculated as follows:
-_n-(+) or
h,
(47)
P-H ,."__Zlr,_
(48)
where
h,:
height from top of foundation to collar (feet),
15. Foundations for Guyed Towers It.was shown by equation (1) that the total soil .toadrng. to be considered in the desisn of tower foundations, is the sum of (S,) whictiis the dead load, and (Sr) which is the toia due to the overturning, or wind moment. In the case of the guyed towers, there is no overturning.-an moment, hoiever, the wind pressure does have imporlant effeci ol the foundation, as the soil is required to resist th.e vertical component of the p,ril on the guy
w1res.
guyed t-owers, therefore, .beFor revised as follows:
, S:S+S"
(49)
wnere
Ss: unit soil loading
equation (1) must
d-ue
to the pull on the guy
(40)
(41) - Rr: Rc csc I ft": pull on .guy wire due to wind pressure (pounds) Ro: horizontal wind reaction at collar (pounds) d: angle that the guy makes with tlie vertical
where
(degrees )
The value of the angle a will usually lie between 30 and ?5 degrees. The vertical component of the pull on the guy wire can be expressed in any o? the follow'in! ways: R" )( cos d (42) R"Xcos0 Si,
R": (R" * Rt) cos o (4s) R",: vertical component of pull on guy wire (pounds) Rt: initial tension on wire (pounds).
wire. (pounds/sq ft.)
Rr:#-
the
where
is
sometimes found desirable to mainiain "stibititv b.y melgq of guy wires rather than a large foundition. Although it is not uncommon to find two or even three sets of guy wires on one stack, towers seldom have more than one set, and even these cases are rare. This discussion, therefore, will be confined to towers with one set of guy wires. Four .guy wires are usually useE ior each set, _ although in some instances three, and in others as many as six have been used. They are attached to a.rigid_ collar which is located it a point approximately 2/3 (sometimes %) of the tower height above the foundation.
-
the.sum of the pull due to wind pressure and initial tension as follows:
a-
RoXcotd
(43) (44)
It is important to note, however, that there is _ always some -initial tension on the guys which must be considered. This initial tensi-on mav be assumed to be 5000 lbs./sq. in., which amounis to 1900 lbs. lor /2" wires and ZSO tbs. for fu,, *ir.s. The weight of the wires may be neglected, when using these figures. The actual vertical component will be a function of the total pull on the guy wire, which is
The value of (S") can be determined as follows:
^R"
ra_
a
(s0)
From equation (2)
^w
a
(2)
Substituting in equation (49)
^
W+R. a
(sl)
16. Foundation Bolts for Guyed Towers The foundation bolts for guyed towers are required to resist the shearing iction of the wind pressure at the base of the tower. It is obvious that ample allowance should be made in the size of the bolts to provide for the initial tension due to tightening the nuts, and also for corrosion. The shear at the base of the tower, which must be resisted by the bolts, is equal to tlre horizontal reaction to the wind pressur-e at that point. This is equal to the difference between the^total wind
l3
at the collar and
can
wind reaction, or shear, at the of the tower. (pounds)
base
pressure and the reaction be expressed as follows:
Rr:P--Ro : horizontal
where Rr
$2)
17. Stress in Shell of Guyed Tower The wind pressure acting on a guyed tower pro-
duces
a negative bending moment at the collar,
has been widety published. There are really. two formulas; one ioi piles driven with a drop hammer, and another for piles driven with a steam hammer, as follows:
For drop hammer o_ 2Wrf . ^ - po+l.o
For steam hammer o_ 2Wrf
'-
where P
: Wh: f: ph:
+-=t=*
i--it+
(ss)
pof
(s6)
0.1
safe load which each pile
(pounds)
will support.
weight of hammer. (pounds) height of harrimer fall. (feet) penetration or sinking rrnder the last blow, on sound wood. (inches)
Care should be exercised
in driving piles,
to
to develop their full strength, but they should not be driven too much, as this practice results in splitting or breakassure that they are deep enough
FIGUR.E 5
and a positive bending moment between the base and the collar. The maximum values of these two moments can be calculated as follows : PM":-lE-(H-h,)'
(s3 )
r\.r,:-!*(r-+)
G4)
ing, and greatly reduces the load carrying capacity. Although piles have been driven with a center to center spacing as small as 2' 6", it- is strongly recommended that this distance be not less thbn 3' 0". Closer spacing disturbs the ground sufficiently to greatly reduce or destroy the frictional resistance.
The top of the piles should always be cut off below the water level, otherwise they will decay rapidly. The reinforced concrete cap is constructed on top of the piles in such a manner that the piles exlend about 6" into the concrete (see Figure 6).
where
Mc: negative bending moment at collar. (foot pounds)
Mn: maximum positive bending moment collar and base. (foot Pounds).
between
Having determined the bending moments, the in a given shell, or the shell thickness required to reiist the bending moment -may be.-cal.iulated by substituting the value of (M") or.(Mr) in place ot 1U,; in equations (32-b) and (33). stress
19. Stresses
in Foundation
After having selected a foundation of such size and shape as- to fulfill the requirements of the problem-from the standpoint of stability and soil ioading, it becomes necessary to calculate the stresses in the foundation itself, to see that they do not exceed the allowab1e limits. The first step in this procedure is to determine
18. Piling
In cases where the safe soil loading is very low, it is sometimes found difficult to design an ordinary foundation which will not overload the soil. In such cases it is desirable to support the load
on piles rather than on the soil. Wooden piles are ordinarily used, and they vary greatly in length, depending on the nature of the soil. The diameter at the lower end is about 6"; and the diameter at the top is about 10" for piles not over 25 feet in length, and 72" for longer piles. Wooden piles generally depend on the frictional resistance of the ground for their load carrying capacity, as they have comparatively little strength as columns. The safe load which a pile will support varies greatly in different localities. Building laws sometimes g'overn the pile loading, and in such cases, the load is usualy about 20 tons per pile, although occasionally 25 tons is permissible. When conditions are not definitely known, however, the only safe procedure . is to drive a few piles for test purposes. The.dommon method of calculating the safe load is by means of what is known as the "Engineering News Formula," which
t4
FIGURE 6
the loading, which consists primarily of the upward reaction of the soil. Figure 3 represents the plan view of a typical (octagonal) foundation, and Figure 3a shows the loading diagram. In this diagram the dead load (Sr) is represented by the rectangle (jklm). The wind load (Sr), which is positive on one side of the centerline, is indicated by the triangle (-p*). On the opposite side of
the centerline the wind load is negative, thereby counteracting a portion of the dead load (wlc). The actrral soil loading will therefore be iep.esented by the area (jkcp). However, the weighi of the bas-e, and of the earth fill above-the base"(area jkno -Figure 3b) do not exert any upward forie on the foundation, and may therefo.e be ded.ucted from.the total load, for the present purpose. The effecttzte upward reaction will then be the area (oc, p) in Figure 3b. 19a. Diagonal Tension
The vertical shear, resulting from the upward reaction of the soil, producei diagonal t6nsion stresses in the foundation. The critical section lies at a distance from the face of the pedestal equal to the effective depth of the base, as indicatecl point (Zr) in Figure 3c. In ottrer words, the 9y foundation tends to break along line (ZZr).'The vertical shear to be resisted is equal to ihe net soil pressure on the part of the foundation outside the critical section. For design purposes, therefore, the load will be _ the area_ (oq.p), (Figure Bc) applied over the area.(a, b, (Figgre 3). Becaui6 of the irregular shape ofIg), the load diagram, its magnitude Ean be more conveniently calculated by breaking it up into its component parts, the totalioad (V")-being the sum of the individual loads, as follows:
port Prism -Rectangular Wedge Wedge pedge. Ivramid Pyramid shape or
o"'lt$r""lif*" a, b, u, i, ar tr g b, fu-, ar br ur tr ?r tr g b, fu].
outline In elevatlon
oqrv oqrv rvp
rvp rvp
as follows:
V" ,b'jdr 'd(80) where fd: unit stress in_ concrete (in diagonal tension) due to vertical shear load. (pounds/sq. in.) V": vertical shear load, outside the criticil section (see Figures 3 and 3c). (pounds) b' : width of critical section which serves to resist
dr
-
diagonal tension stresses (line a, br Figure 3). (inches) ratio of lever arm of resisting couple to depth (dr) (see Table 2). effective depth of base measured from top of base to centerline of reinforcing steel. (inches)
Erample No. 4. Check diagonal tension stresses in the foundation considered in example No. 1 : Figure unit soil loading due to weight of base and earth (see Section 7): Concrete base
Earth fiIl Total
Unit soil loading
Line (gf):67.1'(see Table Line (m, *1 :!:57'. Ltne
(ar b,)
Y 57 : g1
67.1
- --
1).
47.2"
: (b').
: -67.r--2 - 47.2-:9.95" 162 Lrne (a, t,): - 114 -24" -2- Table 2). Factor j: .87 (see Load (m, rl : 4lrf{:5651bs.,/sq. ft. Load (qr) : 565 + 522: 1,087 lbs.,/sq. ft. Line (gtr)
Calculate shear load (V")
47.2'xz"
xrffif_t
g.s5"
x24' x
47.2
xz4
#
xZ#44
9.95 X 238y.24:X2 3)/'144 Total (V")
: : : : :
8,550 lbs. 1,805 935 263
11,558 lbs.
Calculate unit stress in concrete (equation 80)
_ !o:
I 1.558 47
2
x
.87
x 2l :
l3'4 lbs''/sq' in'
This stress is satisfactory, as 40 lbs./sq. in. would be allowed (see Table 2).
oqrv
The unit stress (diagonal tension) resulting from this vertical shear load can be 'determined
j:
(mw): 162 Z :81" do:72". dt :21". dt:162" do:72+21+21:174" Line
of Slab Required for Punching Shear The thickness of the foundation slab (bottom course) must also be sufficient to withstand the tendency to shear along line (Z-Zr), (Figure 3c) 19b. Depth
at the edge of the pedestal. This shearing load may be determined as follows:
S,: S.* S.
(81)
The stresses in this case are not distributed over the foundation area, but are concentrated at the edge of the pedestal. Then Su: total maximum unit shearing load. (lbs. per lineal foot of pedestal perimEter). Sr : unit shearing l,oad due to dead load. (lbs. per lineal foot of pedestal perimeter). Sc: maximum unii shearinf load du6 to overturning_ moment, (lbs. per lineal foot of pedestal perimeter).
The value of (S*) can be determined by adding _ the weight supported by the pedesta[ to th;
weight of the pedestal itself, subtiacting the loadcarrying value of the soil directly under the p.edestal, and dividing the difference by the perimeter of the pedestal base as follows: -
63,000 lbs.
32,700|bs. D57oo
J'
tr..
95.700 lbs : ffi :
633 lbs
/sq' ft'
Total unit dead load (S,) (jm, figure 3c) : 1,155 lbs.,/sq. ft. Unit dead load due to weight of base : 633 and earth fill (jo) Net soil load (om) :-Wlbs./sq. ft. : 803 Maximum unit wind toad (S,), (mp) Maximum effective unit shear load (op) t,325 tbs.,zsq. ft.
where Wr: a, : ^ S.rr: Lr:
W"
+ W. + W,L,
(ap S"rr)
$2) of foundation pedestal (top course). (pounds) plan. area of foundation pedestal. (sq. ft.) weight_
maximum allowable unif soil loading. (pounds,/sq. ft.)
perimeter of foundation pedestal. (feet)
Obviously, if the value of (an S"rr) is equal to
t5
or greater than (W" + W" * Wq),_the value of (Snl becomes zero, and (S.) will then be equal to (So). The'value of (Su) can be determined in a manner somewhat similar to that proposed in Section 12. In that section the overturning load was calculated as a function of the periphery of the foundation bolt
circle, by means of equations (27)
and (35). The bolt circle was assumed to be a hollow iylinder, the wall thickness being infinitely small, as compared with the diameter. In the determination of . the shear at the edge of the foundation pedestal, a similar procedure may be followed, substituting (M1) in place of (Ml), and appropriate values of (I) and (c) in iqui[ior, (ZZ)-, dtpending on the shape of the pedestal. -
Reduced
to their simplest forms, the equations
for the ordinary foundation shapes are as follows: Octagon
-16
Mt
-
Ilexagon
-ro
(83)
.814dp'?
Mt
-
(83a)
.832d,'
Souare
Mr
5r--ffip
(83b)
Circle
Jo-
(83c)
.7g5dn'
In these equations (dn) is the short diameter of
the pedestal (feet).
Once the shearing load (Sr) per foot of pedestal perimeter is knowri, it is a iimple matter to caliulate the unit stress in the concrete, by dividing (Sr) by the effective depth of the base, as follows:
-S, roTt
where
f"'
(84)
stress in concrete base - unit shear. (Pounds/sq. in.)
to
due
punching
Note: The factor 12 is introduced for the purfrom (pounds/lin' foot) to ?pounds/lin. inch) is- unit stress (fn) is in terms oi (pounds/sq. in.). pose of converting (S")
This stress is satisfactory, as 120 pounds/sq. in. is permissible. (See Table 2.) In the case of guyed towers, or stacks, the shear load due to overturning moment (Su) does not apply, but is replaced by
("*,) which is the load due to the pull on the guy wires, as follows:
^Rr S3lguyed)
- S.*
l9c. Reinforcement of Base for Upward Bending Reaction of Soil In designing the base of the foundation to resist the bending moment due to. the upward reaction of the soil, the critical section is located at line (ab), (Figure 3d) along one face of the pedestal (top course). The moments are therefore figured about line (ab), on the basis of the load on the trapezoid (abfg). The load which serves to produce the bending moment in the base is the "unbalanced"
upward reaction. Since the weight of the base, and the weight of the earth filI above the base do not contribute to the bending moment, they may be deducted from the total load when calculating the bending moment. The effective loading will therefore be represented by the area (o qr t, p) Figure 3e. The load is assumed to be applied at its center of gravity, and the moment figured about line (ab). Due to the irregular shape of the load diag'ram, it is difficult to locate the center of gravity, and it is therefore more convenient to break it up into its component parts (prisms, wedges, pyramids, etc.), and figure the moment of each part separately. Obviously, the total moment (Ms) will be the sum of the individual moments. In the case of the rectangular prism, the lever arm used in figuring its moment will be one half of the distance from point (a) to point (t), (Figure 3d). In the case of the wedges and pyramids, the lever arm will be two-thirds of the distance from point (a) to point (t). The individual components and their respective lever arms are as follows: h dor, rlc.3il
Outlltro
Erarnple No. 5. To illustrate the procedure, the punching shear will be calculated for the foundaiion considered in example No. 1. Calculate dead load shear (S.) by equation (82)
W.: W, a,:
30,000 lbs.
ft. X l50lbs. ft, 8:19.87 ft.
119.2 cu.
29.8 sq.
Ln:2.484 \
W. :48,500
: S.rr : -
S.-
:
lbs. 17,850 lbs.
2,000lbs./sq. ft.
(29.8
X
1,850
2,000)
lbs./lin. foot.
Calculate shearing stress due to overturning moment (S,) by equation (83) Mr: 200,(X)0 foot pounds (see section 7)' dn": C - 36, S': 200.m0 :6,820 Pounds/lin. foot
:ffifi0
S',: 1,850* 6,820:8,670 pounds/lin. foot. dr- 21" ttp- 8.670 pounds/sq. in. -::34.4 l?)(zl -
t6
(81)
(84)
(8la)
L
Rectangular
Prism
Outlino ln elevation, Fig.3c
abut
Lever Arm
Distance (at) QrfrVrO
2
Distance (at)
Wedge
atg
QrfrVrO
Wedge
bfu
QrfrVrO
Wedge
abut
frVrp
Pyramid
atg
frVrP
Pyramid
bfu
frVrP
3
Distance (at) 3
Distance (at) 3
Distance (at)
-
\2 XZ
X2 X2
Distance (at) X 2 3
In calculating the amount of reinforcement required. it is asJumed that the portion of the base 'designated (abut), (Figure 3d) acts as a cantilever beafi (of .eitattgula. cross-section) having a width equal io one fa.-ce of the pedestal (ab), a depth eciual to the effective depth of the base (d1) and alength equal to (at). Having ialculated the bending moment as, prostep is to check the depth posed ab-ove, the 19rt
of the base, and determine the amount of reinforcing steel required. These calculations are based on the commonly accepted formulas for reinforced concrete. (It should be noted that for this purpose it is more convenient to figure the momenti in terms of inch-pounds, as the stresses in concrete and steel are usually given as pounds per square inch, whereas in figuring soil lbading foot-pound units are used, as soil loading is uiually itated as pounds per square foot.) For balanced desi(tn, that is, conditions in which both concrete and steel are stressed to their full allowable capacity, the required depth (ds) of the base may be rletermined as follows:
a,:i#k_ where
(8s)
depth of base, measured from top of concrete to centerline of reinforcing steel. (inches) Mo: bending moment in base. linch-pounds) ' f": safe working stress, reinforcing steel in tension. (pounds per sq. in.) (1r
should therefore be placed within the limits of the beam width (ab). However, additional reinforcement should be installed to reinforce the base between the points (gt), and also at (uf),
us.rng_ the same type and spacing of bars as determined for the beam section (ab). This additional reinforcement insures that the entire area of the base is reinforced and weak spots eliminated. Obviously, the reinforcing- bars should extend entirely across the base. Also, there should be a set o{ reinforcing bars parallel to each of the axes, i.e., four sets of bars for an octagonal base, three sets Jor. a hexagon, etc., thus providing strength in all directions. There should be at least 3 inches of concrete below the reinforcing bars at the bottom of the base. Reinforcement in other parts of the foundation should be covered with not less than 2 inches
of concrete.
-
jb,: A":
: ("/",0, ) ratio of effective area of reinforcing steel to effective area of concrete. ratio of lever arm of resisting couple to depth (d').
rvidth of beam (line ab, Figure 3d). (inches)
effective cross sectional area of steel reinforcement in tension. (square inches)
If the design is
balanced, that is, the actual of the base (d1) is that calculated by equation (85), the value of (A") may be determined d-epth
as follows
:
A": b" dr p" (86) If the depth (dr) is greater than required by equation (85), in which case the steel is stressed to its full capacity but the concrete is understressed, the value of (A") becomes : .Mr ,'" (87) - f" jd, . If the depth (ds) is less than required by equation (85), it is recommended that ihe diminsions of the base be changed to give the required depth. In case circumstances make it impossible to increase (dt) to the required dimension, it will be
E.rample No. 6. Determine bottom reinforcement for the foundation referred to in example No. 1. Figure bending loads
Line(m,w):,72" -36" : 357 pounds/sq. ft. Load (m,.,, :!9i*4 Load (q, r,) : 357 + 522 - 879 pounds,/sq. ft. Load (v, p) : 803 357 :446 pounds,/sq. ft. Figure moment (M5) Line (ab) :29.8" Line (ta) :45" , Line (gt) : 18.65" 29.8',Y
-i?4
18.65
X 446X
45
144X3 Total (Mo)
'/)v
/\e/\
ft'
45"
X 2
45X2 3
:184,000in'-lbs
:
153,000
:
62,300
:
51,900.
:451,200 in.-lbs.
Check depth of base for balanced design (equa-
tion
(85)
: : j: b": f"
p.
necessary to increase the amount of reinforcement
used. The determination of the amount of reinforcement required for sdch special cases is beyond the scope of this article, and reference is made to the various publications dealing specifically with concrete design for further detaili. Having calculated the cross sectional area of steel required, a selection is made as to the diameter,.shape, number and spacing of bars which will give the required area. It ii recommended that the center-to-center distance be about 4 inches if possible, but not less than 2l times the bar diameter for round bars, or 3 times the side dimension for square bars. Generally speaking, a large number of small bars (1, s/s, o, % inch)-are preferable to a smaller number of larger bars. It should be borne in mind thaf the area of reinforcement determined above is the amount required for that portion of the foundation having a width equal to (ab), Figure 3d, which was assumed to be the cantilever beam carrying the entire bending load. This amount of reinforcement
45"
tql" X 879 Pounds'/sq' 18.65 X 45 45 X 2 - u4 x87ex--29.8 X 45 446 45 X 2 14442/\3
18,000 .0089 '87
29.8"
f"P"j:138.7 I - 4il:oo :10.5" 0t(hnrlnccd' : Y t.lg-z x, zg: Since the actual depth is 21 inches, whereas only 10.5 inches would be required, the concrete will be understressed, and the area of reinforcing steel should be calculated by equation (87). A"
:
451.200 .87
tsPoo
x
x 2l -
1 37
sq' in'
lJse )1 inch deformed square bars (.25 sq. in.
area).
Number required #+ :5.5. Use 6 bars within the width of beam (ab). 29.8" :4.96". l;se S-inch spacing,-enSpacing ? -tid" tirely acros. (gf), which will require {11 :13 bars per set. Four sets of bottom reinforcing bars will be required for the octagonal foundation.
t7
to Resist Stresses Due to UPlift As explained previously, the wind moment creates a positive sbil load on one side of the centerfi""-, utia i negative load on the opposite side' In oi-frlr-*otas, ihe action of the wind tends to lift tt fo""a"tibn on the negative side. This upward " oi "uplift" effect, G resisted by the -*gigh! i;;;; of ;i iht "o""i"t" base itself, and byItthe weight-betherefore ihe- earth fill on top of the base. cornes necessary to reinforce the top,of the base, resist the reiulting negative bending moment' to --tfr" pro"edure is qiit" ii-itur to thaf described soil'reaction (Section 1.9c)' The foi1t "'rp*ard (abfg), and the outline of the area the in ioua r.tt itt"o."tiiut beam carrying ihe load is (abut) as in S..lio" 19c. Howevei, i., this case the load is the *Li"t,t Der square foot of concrete base, plus the *"iilt i i"r tor"t" foot of the earth fill, and is unithe calculair#it iistrituted, thus simplifyingthe reinforcemoments, the ii;;:'Aiter figuring ment is determined 1n exactly the same manner as exptained in Section 19c, using the equation
Moment
19d. Reinforcem€nt
^-Mo f" jdt ^'-
2%!L;570X
E%q
:
4+Z
: :
Total (M")
119,000
in.-lbs.
99,500 218,500
in.lbs.
From equation (87a) 218,500 inch lbs.
in. within beam A":-l8p00X.87X22 -.636 sq.(29.8") width
lJse rft-inch deformed square bars,
at
10-inch
centers. 19e. Bond
In order for the reinforcement to be effective, the'-st-rength of the bond between concrete and be sufficient to permit the reinforce;;i;;t? stress *""1 t" develop its full strength. The bond following the of meais by ;;;- t" ialcul'ated formula:
V,
U: EJA; where ""-';:
(87a)
(88)
bond stress per unit of area of surface of bar' (Pounds)
>": 6erimeters of bars -within the limits or - ;;;;f it. beam width (ab)' (inches) Erample No. 8. Check bond stresses in example
In this case, (ds) is the depth of the base from the centerline'oi the upper layer of reinforcement ;;ih;-i;;i;* of the bise, and (M-) !s the bending (inch--pounds)' forces uplift the to a"" *o*.ttt Eramble No. 7. Determine top reinforcement to r.Sri "'pfitt in the foundation ieferred to in example No. 1. Weieht of concrete : 300lbs./sq. ft. 150 lbs./cu. ft.:r.Ztt. Weieht of earth :27Olbs./sq. ft. 60lbs.,/cu. ft. X 3 ft.
No. L. Bottom reinforcement V.
11.578
>.:6
lbs. (See examPle No' 4)
X.5 X 4:12" By equation (88)
u:-nffil:53 11.578
:T-701bs.,/sq. ft.
Totat
y57oX
+-
lbs.
Bond stress for bottom reinforcement is satisis permissible (see Table 2)'
fr.tlil "t ts por"as Top reinforcement IABLE 2
GonrlonE Apptylng
io
Figure shear
gz#y-
Foundotlon Dcslgn
Mlxture: Cement'
2 5
2 4
Sand. . .
CoarEeAEgreEat€''""' fb
Safe
baring load on concrete (lbs./sq' in') '
500
375
fi
Ultimate compressiv€ strength (lbs'/sq' in')
2.0(n
1.500
f.
Safe --ii. unit stress in
800
600
fa
unit stress in concr€te due to vertical Safe --"i".i-fai.co"al
ir).
40
30
fD
Safe
unit strss in concrete base due to nunchinc sher. (lbs./sq. in.)...'..'..
120
90
f.
Safe working stress, steel reinforcement in tedsion. (lbs./sq. in). . " '
18,000
18,000
i)
(inch-pounds).
15,600
16,000
ir,
(inch-pounds)
138.7
88.9
i
Ratio, lever arm
.87
.89
15
l5
(f.
(f.
qtreme fiber of concrete ompression)-(lbs./sq. in').." " "' tension) (lbs./sq'
F%J {"=oikJ *u
.
'
.
depth (dr).
of
resisting couple to
Ratio, modulus of elasticity of steel to that of concrete,
Ratio. efrective uea of tension reinforce ment to efiEctive ilea of concrete ' ' ' ' Safe bond stres --i6rje.&ti
(concrete
of surfae 45
m
* These figures may be slightly incresed by making "U"-bends on the
""a";l;|iif,l.i",i[f tipe-. riri tionJ
iiijre
of intqest.
t8
as most satisracrory ror rounda*?Ii;," is recommend_edniiture are presented as a matter ionstantJ ttiiiiii ilztr
:
947
:5,427 lbs.
Total (V.)
s427 $:- 6ffikT
(88)
-48lbs' The bond stress in top reinforcement is satis-
factory, as ?5 pounds would be allowed. 19f. Bearing Stresses
The bearing stresses (where th.e steel tower
rests on the cincrete pedestal) seldom cause any preaif1."ltu, but should lre checked as a -safetystress .rrlio".'tfte bearing stresses consist of due .the to the a""-1o *i"d pressuie, plus the stress dead load as follows: Bearing
60 75
:4,480 lbs.
57orbs.
2o:3X.5X4:6
.0056
to steel rein-
*i irnit of arm
of bar. (pounds)
.0089
X
2W-xsto
I
I
stress:
4M*/7tD"
+ (w'* W")'/z'D'
(37a)
(See Sections 11 and 12.)
ilquation (3?a) gives the bearing stress in po""tt p.t ii.,eil Toot of shell circumference'
These stresses are spread over the are-a of the base purposes the unit bear;"". tfr.tfore for prictical -determined as follows: i.,gjsttess can be
fr:
4M. W, + w" 17+ ?r D,
,r
(37b)
l2r*
in which
r,: width of the tower base ring. (inches) fb: unit compression stress on concrete. (pounds/sq. in.)
Equation (37b) may be modified slightly, depending on the exact shape and arrangement of ihe base ring (or base plate), but in the majority
this case are given for illustration only, the design has not been changed to take maximum advantage of the allowable stresses. The stresses in foundations of this type should not exceed those commonly accepted as good engineering practice in reinforced-concrete design, for the particular mixture of concrete used. As a matter of convenience Table 2 is presented to show allowable stresses and miscellaneous constants applying to two grades of concrete quite generally used for foundations. It is strongly recommended that the 1 :2 :4 mixture be used in practice, the figures for the 1 :2:5 mixture being shown primarily as a matter of interest. 19h. Suggestions and Recommendations
TOP
The calculations explained above provide for reinforcement to resist the stresses due to the
OF
PEDESTAL
various types of loading. It is good practice, however, to install additional steel as a means of tying the foundation together, to form an integral unit. The same size bars are used for this purpose as for the main slab reinforcement, and the designer must use his own judgment as to the number and location of the bars. Figure 4 represents what is considered good practice, and is offered
4.F4 ''a 'c .
.,SLEEVE NUT
'.4.'..y.4
4.4 " :.WELD
as a guide. In the case of very large foundations, consider-
FIGURE 7
cases it may be used reasonable accuracy.
of
in the
above form with
For guyed towers, equation (37b) becomes: 4M,
fr:
r
,r DF 1-
R"+W"+W" z'D,
(37c)
19g. Allowable Stresses in Foundation It is to be noted that in actual practice the depth of the base in the examples given above could be reduced, if desired. All of the stresses for diagonal
tension, punching shear, bending (upward ancl downward) and b-ond in the reinforcement are well below the-allowable values. As the examples in
able concrete and weight may be saved by constructing the pedestal with a hollow center, as illustrated in Figure 5. Of course, the inside form is left in place. It should be noted that the base slab extends all the way across, to provide protection and bond for the reinforcing bars. Foundations supported on piles should be so constructed as to allow the tops of the piles to extend about 6 inches into the base, with the bottom reinforcement about 2 inches above the piles. (See Figure 6.) Considerable inconvenience is sometimes encountered in setting the tower in place, due to the difficulty of lowering the heavy vessel over the foundation bolts without bending some of them or damaging the threads. Figure 7 illustrates a method of overcoming this difficulty. A sleeve nut is welded to the top of the bolt, and so placed that the top of the nut lies slightly below the surface of the concrete, with a sheet metal sleeve around it. The tower may then be placed in position without interference from the bolts. Stud bolts are next inserted through the lugs on the tower, and screwecl into sleeve nuts from the top.
Nomencloture Ar
: elfective cross sectional area of sion (square inches)
steel rcinforcement
ir
ten'
For balanced design (86) Ar: b" dr pr If depth (dr) is greater than required by equation (85) e"
:--14.0.
(87)
tr ldr For top reinlorcement of slab to resist uplift strcsses:
o.: ts-r1,, Jdt
c
c'
:
(87a)
face
assumed to dct as a cantilever beam resisting the'bendini stresses (line ab, Figure 3d) (inches)
section which serves to resist the diagona! tension stresses. (line ar br, Figure 3.) (inches) distance from neutral axis of foundation base to point of maximum stress. (feet) distauce from centroidal axis of foundation base to any point under consideratiofl. (feet) outside diameter of tower. (feet) inside diameter of tower. (feet) tower diameter measured over insulation. (feet) diameter oI Ioundation bolt circle. (ieet) short diarpeter of foundation base. (feet)
D: : Do: Du : dr : dc : short diameter of critical section for diagonal Dr
a : area of base of foundation (sq, It.) ar: plan area of foundation pedestal (sq. ft.) B : barometric pressure (inches Hg) ba: width of the critical section (equal to the width of thc
of the oedestal)
b'- width of critical
dr
:
tension
stresses (see Figures 3 and 3c). (inches) effective depth of base of foundation, measured from top of base to centerline of reinforcing steel. (inches)
t9
dp : short diameter of foundation pedestal (feet) Ec : modulus oI elasticity of concrete.
E! : e
modulus of elasticity o{ reinforciflg steel. eccentricitY. (feet) This iactor ia the distance from the centroidal axis of the foundation to the point at which the resultant of the dead load and the rvind ioad intersects the base of the foundation. The eccentricity can be calculated as follows I
tr{r
(12)
Eouation (12) sives the eccentricitv at the condition of oobrest stabilirv, that i", with the miiimum dead load. This is the value which ordinarily is used for design purDoses, for maximum however. it is obvious that the eccentricity''ubstituting the a""a toia ".nditions ian be calculated by (Wt). The plaCe (W) (12) in of in equation value of marinum valtie rvhich it is possible for- (e) to have and still maintain the stability of the foundation is t2
em.x: emir :
(
Or
1e)
(19a)
-,
Values - of (e-"") for various foundation shapes are follorvs:
Octagon: ens:
Hexagon; em.':
Square: Circle:
0.122d 0.121d
: 0.1 18d em!:: .l-d emax
as
f"' : i" : fp
:
,,,
relation to the moment of inertia can be expressed as follows;
I: r
in. )
H: height of tower. (feet) hr: height of foundation. (feet) hr : height oI collar (to which the guy wires are attached) above
(68) 7.
k for various foundation o ? )-
";-'
: *'1'" Circle: ,.: *1'"
Square: L:
lever arm of
I:h,f
shapes are as follows
a 1<-
k: -J:fr:
Ilexagon: k: u
(72) (72a) (72b) (72c)
wird load (feet) to be calculated as follorvs: lt (r)
Sr: unit soil loading due to dead load. (pounds/sq. ft,) Srh: ulit soil loading due to minimum dead load (pounds/sq. ft.) to include the rveight of the empty tou'er, the loundation and the earth 6ll only. It does not include insulation, platforms, piping, liquid, etc.
Sz: unit soil loading due to oyerturning moment. (pounds/sq. ft.) Sa: total maximum unit shearing load, (pounds per lineal foot of pedestal perimeter)
Sr:unit oi
Mr: bending moment in base. (inch-pounds) Mc : iegative bending monrent at collar. (Ioot-pounds) (see equation 53) Mr : overturning monetrt about base o{ foundation. ({oot-pouflds) (7) NIr: P* L tr{p: maximum positive bending momeilt between collar and base. (foot pounds) (see equation 54) bending moment about base of tou'er, (foot'pounds) (;
Nr,: P*
(+)
(26)
: bending moment in base, due to uplift forces. (inch-pounds) N : number of foundation bolts q : (E"/Ec) : ratio, modulus of elasticity of steel to that oi concfete. P : safe load which each pile will support. (pounds)
Mu
20
Sr:
Sa
* Sr
(81)
to dead load. (pounds per lineal foot
shearing load due pedestal perimeter)
* \Vs+.W"+Wp-(apSr:r') Lp
-{
So: maxinrurr unit shearing load due to overturning (pounds/lineal foot of pedestal perimeter).
(82 )
moment.
For pedestals of various shapes, the values of (Sr) are
as
follorvs;
Uctagon: 5s: llexagon' Square: Lrrcle:
Sr
:
IIr
(83 )
j14dr" Mr
(83a)
trfr
(83b)
: -gJZdt .nag:p ^Mr 5r:: so
.
/
6)d,p
(
83c)
srrr : maximum allowable unit soil loading. (pounds/sq. ft.) Su : load on each foundation bolt. (pounds) Sg: unit soil loading due to pull on guy wires. (pounds/sq. ft.) Smrn: total unit soil loading under minimum dead load conidtions.
St: t: uV: Vs : W
:
W":
Lp:perimeter oi {orrl"tior pedestal. (feet)
Mt:
(59)
(s7)
Sz, also
Octagon,
(l)
S:S'(1+k)
{ollows:
Values of
(2s)
r* : rvidth o{ tower base ring. (inches) S: total unit soil loading. (pounds/sq. ft.) S: Sr * Su also,
foundation. (feet)
: .ea
(17)
t:1; ll
f : moment of inertia of the base of the loundation. (based on dimensions in feet) j : ratio of lever arm of resisting couple to dePth dr. (See table 2) k : Iactor bv which the soil loading due to dead load must be -uiiiptiia to equal the soil loa-ding due to overtutning, as kSr
ar'
earranling:
(22c)
fr : safe working stress, steel reinforcement in tension' (pounds,/ sq.
'":(A"/u"a')
Rr: horizontal wind reaction or shear at base of tower, (poutrds) : horizontal rvind reaction at collar. (pounds) Rs: pull on guy rvire due to rvintl pressure. (pounds) _R" (40) I(s: S.n t or, (41) Rg:cscd Rv : vertical component of pull on guy wire. (pounds) (45 ) R" : (Rs f Rt) cos d Rt : initial tension on guy wire, (pounds) r: radius of gyration of the base of the Ioundation (leet). Its Rc
(81 )
- t2d,
torver.
concrete,
(22a)
in. )
(8)
pb: penetration or sinking oI pile under the last hammer blow, on sound rvood. (inches) pr : ratio, effective area of reinforcing steel to effective area of
(22b)
barometric pressure. (inches Hg) height of hammer fall. (feet) unit bearing stress on concrete. (pounds,/sq in ) (See equa' tions 37b and 37c) safe unit stress in extreme {iber of coucrete (in compres' sion). (pounds/sq' in.) in ) ultimate compressive strength oI c-oncrete .(nounds,/sq iiiii it*.i in' concrete (in"diagonal tension) due to vertical shear load. (Pounds/sq. in.) unit stress in concrete base due to punching shcar' (pounds,/ sq.
(pounds per sq. ft.)
(22)
The value of (e) as calculated by equation (12), and based on the minimum dead load (Wt) should ,r.i'.,'exceed the value calculated by equations (19) or (l9a).
F: f: fu: fc :
Pw: total wind load (pounds) to be calculated as followsl P*. : pc Do H p: wind pressure on a llat surface. (pounds per sq' ft.) Dc : wind pressure on projected area of a cylindrical
1\'r: Wp: W": Wt:
(pounds/sq. ft.)
unit stress in tower shell due to bending moment (Mt). (pounds/sq. ft.)
shell thickness. (feet) bond stress (between concrete and reinforcing steel) per unit of area oI surlace of bar, (pounds) velocity of wind. (miles per hour) vertical shear load outside the critical section (see Figures 3 and 3c). (pounds) total weigtrt on soil (pounds) calculated by' the iollou'ing
equation:
W:Wt *
W.
(3)
of auxiliary material and equipment supported by the foundation (pounds), including liquid in the tower, insulation, platforms, piping, etc. (Does not include weight ol rveight
tower. )
rvoiglrt of hanrmer. (pounds) *"1*1r, of foundation pedestal (top course). (pounds) rveight of empty tower. (pounds) nrinimum dead load .(pounds) rvhich is the yeight of the einpty to*'er plus the weight of the foundation, inLluding the earth 6l! ou top oI the base. Z : section rnodulus of the base of the foundation (to be based un f,,urrdatiorr tlimensions in feet) as follous: Z
:.1
(11) c
r : 3.1416 d: angle rvhich guy uire makes with the vertical. (degrees) f,o: sum of perimeters of reinforcing bars within the limits the beam width (ab). (inches).
of
Simplified Design for Tower Foundations Curves reduce design time for octagonal reinforced concrete tower foundations by quick selection of base size, thickness, reinforcement area and unit bond stresses
eao 5a 'ie oC sO E= o,OLL
Andrew A. Brown, Union Carbide Chemicals Co. South Charleston, W. Va. DBsrcNBns oF FouNDATroNs have used many different locations for sections and beam widths to compute bond
shear, bending moments, and diagonal tension shear. Since agreement on these important phases is not complete, this presentation uses The American Concrete Institute Building Code Requirements as a guide for
reinforced concrete design and the allowable unit stresses therein. The usual assumptions are followed as to the behavior of reinforced concrete and soils. For simplicitn the derivations of fornulas are based. on the inscribed circles of the octagonal base. This does not influence the accuracy of the final results. The foundation engineer is ever mindful of the fact that a substructure design based on inexact soil bearing determinations, concrete with variable strength, and loads which can be off 10 percent or more, is not very definite. The application of good judgment coupled with experience is more important than carrying out computations to more significant places than the informatior and assumptions
waftant.
Foundqtion Size. As the size of the foundation is the first_design requirement after the permissible soil bearing has been established, the formulas used for this determination will be derived in that order. When the resultant of all forces acting on the foundation strike the base within the kern, the forces acting on the. soil can be represented graphically by a right .eg.rlu. cylinder resting on an ungula of a right regulir cylinder. If it is on the edge of the kern the soil reaction forces form an ungula whose base is a circle; when it is outside the kern, the ungula has a base in the form of a circular segment (Figure 1). The volumes of these solids are equal to the total weight supported by the soil, and their moments about the center of the base are equal to the moments of the external forces acting on the foundation about the same place. Then the eccentricity ,,e,, measured from the center of the foundation equals external moment of ail forcep (M) Total vertical Ioad (W)
which is equal to the moment of the forces acting on the
bottom of the foundation divided by the total forces acting on the base.
Resultant Within the Kern. For the condition where the resultant is within the kern (the area inscribed by a
lnscri bed Circle
Equ i vo lenl Squo re
ACI
1208
Section for Bond ond Mornent ACI 1204 (o),1205(c)
FIGURE
l-The
radius equal to
resultant of all forces is within the kern.
/s of. the diameter of the circular founda-
tion) the maximum soil pressure P is equal to the total height of the right circuiar cylinder and ungula drawn to graphically represent forces acting on the base. For this condition the maximum soil bearing
* (1+g\ ,D_- 7R,\ D/ and the minimum soil bearing equals the height of the soil pressure cylinder or
w (1t: z.Rr\
B"
\
D/
Let V equal volume of cylinder and ungala
which
equals W, the total vertical load. To get the general formula, let the maximum soil bearing equal unity and h equal minimum soil bearing, then
the total load
w
:
7R2h
For a value of
*
rz'Rz (1
|lers D
-
h)
than r/s, the maximum soil bear-
ing (unity) can be computed in terms W and D. As an example, for
elD: .tl,pr: *
(r +
w (1-.8): A
.B) and p*,o.
:
.2W
i
2t
SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS
If
unity or P,
1.8 is reduced to
then P,,ir.: f
..
'
, tn"
height of the right cylinder and the height of the ungala becomes
8/e.
Then
.irpz/
1 \ *:-i(\r+L.o)-9-orcuR:
5zgz
C, is a coefficient which when multiplied by the product of the maximum soil pressure and radius squaied will give W, or the total volume of the cylinder Where
Resultant Outside the Kern. For the development of the equations for moments and total forces acting on the base when the resultant force is outside the kern, refer to Figures I and 2, which show this condition' To get the volume of the ungula of height P, whose base is bounded by the angle -+ d as measured from the X axis, we have dV : d A P'. dV is a volume whose area of base is dA and height P' and is located a distance R"o" { from the Y-Y axis. Then by similar triangles pr
- -
rP-r (Cos \v"." d Cos --- a)/
(1-Cosa)
:
dA:2R
Sin
{
dx and
dx:R Sinddd.
and ungula for this condition. For the values of f, , the coeflicients C* were computed, Column 8 Table 1, and C values, Column 9, were obtained by dividing C,R'!by
/e\
* (;/' fABLE l-Coeftlcients
for Yorlous e/d
v
e/D
e/D
M/2v .10 .15 .20 .25 .30 .35 .40 .45 .50 .55
.60 .oD
.70 .75
.80 .85 .90
.95 1.00
.0603 .1045 .1516 .1988
.457 .436
.079
.415
.2E4
.395
.2448 .2879 .3275
.374
78'28',
.0660 .1198 .1823 .2518 .3269 .4068 .4S04
-403 .583 .811 1.099
16',
.5773
90' 96" 44', r01'32',
.6666
.3626 .3927
.Bt4 .294 .275
36'52', 45" 34',
53'08', 60' 66" 25',
84'
107'28', 113" 35', 120"
126" 134" 143" 154" 1 80"
62', 26',
08',
09'
.4172 .4354
.7679 .8504 .9440 1.038 1.128
.4493 ..4567 .4557 .4537 .4408 .4297
t.226 1.312 1.404 1.488
t.57t
.354
.256 .238
.220 .202
Volues
.t67
t.463 1.92t 2.505
CvB2
.12 | .11 |
.lo I .os I .08 |
Y /4(e/d\2
1.602 *l 27.82 1.6711 34.52 r.zqsl le.ns 1.8251 56.34 t.grol zl.s,l
.oz I 2.013 1102.7 .06 I z.tzt 1117.3 .os I 2.244 1224.4 .04 i
.035
|
2.880 1371.8 2.454 l5oo.8
3.235
4.170 5.370
.4t07
.185 .168 .153 .138
6.97 8.S4 11.65 14.99 19.60
.3927
.125
ao.tn
Reultant inside kern
FIGURE 2-The resultant of all forces is outside the kern'
Resultant outside or on edge of kern
c t000
.r0
2?5 .250
.225 .200
.20
.30 .40 .50 .60 70.80901.0
20. r3.5
r4.75
FIGURE 3-Curve used to determine soil bearing or diameter of foundation
22
base'
Deduclion for Fill ond Slob | .s0
.85 .80
I
l.ao
.75
l.ru
-70
.70
,65
.65 .60 .so .55 .55
of
.50
Ungulo
.50 .45 .45
o to
.40.40
450 ACt
U)
for
Sheor
t205 (b)
o o
35
R(.707-cos4 FIG-U_RE 5-The soil reaction is the sum of the forces shaded area.
or M .oo L
.20
.to
.25
:
CRBP, where C
r:
30
in the
(l Cos a) --1-(2)
By use of equations I and. Z, Columns Z, 4, 5 and. 6 _ of Table 1 were computed for values of K or angle alpha.
.1y substitution,
dv and
: -aa-2R9P. co;)
V
(cos d
:
-
cos a )sin:t'dP
a*rro1co, 2R2P1 [l'*,l - (1 --Cosa)f -cosa (+-+si^cc.,d
+
)]"
2.--_J.
,,,
The.moment o{ any ungula which represents the forces apptred on the base of the foundation about the y_y axis is the summation of the product of th" differential
volumes, dV and R Cos
dM -
So
"o
:
{.
2R3P1
(1
_-Cos;;
[(Cos
d-Cosa)
Cos CSin'z
ddd]
:11--2R3P- la d sin2 d c* O) - cos a cos c sin2 d) dc ","*' 3R'3 . [.---fB /Lr,"or-r\-cosasin,cla (1-Cosa) ^ \4""'aY-Y)--3
l"
2R3P. (1
-
C"-", a f Cos a Sin a-
: w
2Cos3 a Sin a
Cos a $i1s
3J
-
al
I
:
yeight of foundation and equipment and .eccentricity caused by wind ,.ro*"rrt, seismic torces, " " etc.
:
and
C : a numerical
coefficient
value.
for the
respective e/D
With the maximum soil bearing given
^_
M
|
tion can be obtained for a permissible soil b"a.irg o,
conversely, the soil bearing can be computed for a kn"own foundation. The formula for soil bearing is
C", where W
f Sin3a , Sina665za-2e65al
\D/
_ Th: curves orr Figure 3 were plotted by using Columns 5, 6, 7 and g. From these .rrru", the size of tie founda_
P,
Or V (W) - CRzpl where C:
ft-cos"ll 3-]r--
+andColumn6:,V, (z\''.-
cos a) sin2 ddd
c
2
:#:
Column5
w
" - Pr"' which
locater* :
C andD
The relation between
K
and
: f
+. is sho*r, by the curve
on Figure 4 which was formed by using Columns
of Table
i
and 5
1.
Foundqtion Thickness. After getting the size of the is to determine its thickness. Since the missible maximum unit shear is 75 psi this is usu_
base, the next step
23
SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS .
O
*,orn of
Sheor
Footing Resisting
: C r R (See Toble 2 Col.l3).
.25 .30
@ ,0,r."
uo,rr. ol slress Prism lyhose Bose is o Tropezoid ond Segmenl oI Circle . CP5R2(See lobte 2 Col. 3) @
.35 30 .45.5055.@ .70 .80 90
3.0
1.0
3.5
of stress
40 45
Prism
5.0 60 70 80 90
FIGURE 6-Curves for computing shear for diagonal tension.
ally the controlling factor. In many designs this limits the strength of the concrete to 2,500 psi for the most economical foundation. The soil reaction considered in
By integrating and substituting the values of the trigonometric functions for the 45 degree angle, the force V : , , '*'="t . [.1 (l-Cosa)' This formula provided the value in Column 4 Table for the various values of K (a).
height
(.707-
of -]1
The volume of the wedge is R2P4(.707
and
Cos a) P.
Ps. This
When K is .1465 or less, this area is a segemnt of a circle. The volume of the section of the ungula can be solved by application of limits of 45 degrees for { in computing the volume.
(l
P-
-6;;d) TABTE
I
2
)"ttt"'
c cos d-Qe5 a sin'z P)dd
3
4
5
CPsRz
v
Sec, Ung. C'PnR2
Wedge C/PrR2
36" 52', 45" 34'
.1635
:::::
60'
.5354
.0660 .1198 .1611 .1860
66" ?,'',
.6254
.2029
.6954
.74ru
.2144
780 28
.7754
.2302
530 08'
84' 90"
16',
.2954 .4254
.7854
.0193 .0547 .0951
.7726 .2067 .2357
Diagona Tension
24
respective
K
and 4 values the volumes are
!-Vqlus5 to Colculote Diogonol Tension, Bond, Moment ond Beqm width
d
K ,10 .15 .20 .25 .30 .35 .40 .45 .50
(4
Cos 4)2(2 Cos a + 2.828) - 6(1-Cosa)
Bond-Bending Momeni. The slab is now investigated to determine the area of reinforcement and unit bond stresses. The moment of all forces to the right of Section A-A, Figure 1, determines the area of steel, and the sum of these forces is the shear used in computing the bond. Section A-A is located by passing a vertical plane through the foundation along the side of the equivalent square. The external forces acting on the base can be conceived
curve 2, Figure 6.)
2R2
:
These two columns are added (See Col. 6) and the results are plotted producing Curve 3 of Figure 6. The width of the footing "b" for diagonal tension is 2k). When a is 45 degrees or less, it is 2R Sin a. 2R (1 These -values form Curve I of Figure 6 and are tabulated in Table 2, Column 13.
f,t I latter area tr The r'orce is R2Po * -(1-2K')')R2. L f,7r I , and is Column 3 of Table 2. (See L * -f -2Kr)z_1
v:
for the
6
Col.
7
E
I Mom.
IO
una. I Seg.tr,t.CyL
Ungula
Seg, Cyl. CPsR2
CPzRs I
.0660 .1198 .1804 .2407 .2980 .3497 .3958 .4369
.0668 .1198
.1035
-0203 I
.47&
CPrR2
.2955 .4473 .6142
.7823 .2518 .3269
.7527 .9780
.4068
7.773
.490l{ .5773
1.389
1.577
.6666
Bond
.0076 .0422 .oz2s .1140 .1658 .2294 .8049 .3s22
1t
t2
2Rslnc CR
Octag,on
I
4+5
CPrR2
2
recorded under Column 5, Table 2.
and a force solid whose area cos aof base is the area of the trapezoid abcd and the segrnent
of the circle whose chord is cd and of height
.1427 cos a]
I 7g5
computing diagonal tension is the sum of the forces acting between 90 degree radial lines drawn from the center of the base through the two corners of the equivalent square and bounded by section B-8. This section is parallel to the side of the square at a distance "d," (depth of concrete) from it. One can see by Figure 5 that these forces can be represented by a section of an ungula whose height is Pa, a wedge whose base is a trapezoid abcd and
I .0132 -0961 I .oz2g I .12s9 I .rsor I .2829 I Bsz4 Ls195 I .6666
Bending Mom.
1.20 1.40 1.60 1.73 1.83 1.91 1.96 1.99 2.OO
13
I "b" CRICR
1.28 I 1-43 I r.68 I 1.83 I 2.00 I 2.oo I 2.oo I 2.00 I 2.00 I Width Beam
1.20 1.40 1.20 1.oo .80 .60 .40 .20 .00
@ro,ur. of
Stress prism Whose Bose
is
@ ,or.nt :t"_tlT.r A-A,Fis.,=jl;]:-
Segment of
Circle r used in Computing Sheor Cp3R2 =
@ Votr..
of Stress prism
in computing sr,.o,
=
Which is
c16R'
@ *,0,n of Fooling ot Section
Eendirq Momenl =
on Ungut0 ,
used
Moment
O
About secrion A-A used
lor
of
Prism (UngutolAbout Section
Stress prism, Section
o-o
ot
Cylindar
= '18t"
Sheor ond
CR
30 .3s .40.45.50.55.60 :o.ao .so to
30 3.5 404550 6.0 z0 8090
FIGURE 7-Curves for computing shear for bond a,,d bending moment for reinforcement. as being in the shape of an ungula of height pz and. a segmeqt of a right circular cylinder of height p3. The sum of the two volumes is the shear force, and t-he sum of their moments about A-A is the bending moment that determines the reinforcement. The weighis of the con_ crete slab and earthen fill are deducted irom the vertical forces. This is easily accomplished by reducing the inten_ sity of the uniform bearing load acting on the bottom of the base.
The volumes for the ungula have been computed earlier for obtfiing soil bearing and those values for K equal to or less than .5 are shown in Column 7 of Table 2 and Curve B of Figure 7. (See Equation i for V.) The volume of the segment of the cylinder is equal to the product of the area of the segmenl and p.. The area is easily computed by making use of the fact that the middle ordinate is KD. Values for the respective K" are shown in Column 8, Table Z and. plottei as Curve A,
Figure
7.
. T!" bending moment equation is derived by substitut_ ing R (Cos f Cos a) for R Cos { in the development of the formula- for moment about the center of the foun-
dation.
This gives
Iut:
for the 1gs_
The moment on the forces whose configuration is a a cylinder (see Figures t and 2) is derived
segment of as follows: dM
:
/a
2RrPs | {Cor
)"
O
-
:2RBPg l- sirr 6
L--s--_-
:2RsPa l- sin,
o
L--.-
+
Cos
a) Sin2 d d C
Cosa(p-+
sin 2 c)
SinaCosza_aCosa
]"
l
The values obtained for the angles a (K) are noted in Column 10 and form Curve E, Figure 7. The widths of the foundation at the sections are equal to 2R Sin a and. are shown in Column 11, Table 2, and
Curve C, Figure 7. Column 72 and. the dotted curve (Figure 8) indicate the width of beams for any octagon. Use
of Curves. As an illustration of application of
the
th-" following information is given: height of ves"l*9r: sel, i12 feet; diameter, I feet; the anchor Lolt circle
fj*
|""r"*d_-cosoj,si.,,p6p
2Rsp1 [ r /r \ 2cosaSinsd n-c;;iL-- +(it'" 46-o)-
-cos2"(+--.Is;"r-)]: *:
in this equation the trigonometric val_ respective angles cor.esponding to the Ks, Column 9, was obtained and Curve D plottej on Figure 7. By- substituting
2R'.P,
,, , [: r, * (1-Cos") LB'
$"",aSinsa-r*#-]
4 cosz a;
a l0-foot octagonal pier; top of pier is one foot above grade and 6 feet, 6 inches above the bottom of the foundation; permissible soil bearing 5,000 psf (pr) at 5 feet, 6 inches below grade; wind pr"rrr.", Sb pri oi horizontal projection of the vessel. requires
Operating weight of vessel, 200 kips; vessel empty, 100 -. kips; and test weight, 300 kips.
- asuae653a4
The diameter of the .base required under operating conditions will be determined first. The moment of wind force about the bottom of the foundation is 112 x.03 x B x 62.5 : 1680 krp_feet.
2s
..
SIMPLIFIED DESIGN FOR TOWER FOUNDATIONS
.
Estimate the weight of the foundation using a 22 foot, 6 inch octagon, two feet thick.
: Bl kips : 101 : 118 336.2 (.35) Total :300 : tg6 (operating) Weight of vessel W:SO6 (82.8) (6.5) (.15) (419 82.8) (.3)
Pier: Slab: FilI:
EccentricitY
:,|| 680 : 1
"
3.36 feet, e2
11.0
5.?2
-
5.28
-
.153
:
:
11.3
The weight of concrete and fill becomes 280 kips and 480 kips. To compute the maximum unit bearing
1-
^P.
480.000 '""i,
f""t# :
+: :
used above,
C
, :2,900psf (3,000'
r3.5(3.5),
t : *=frffi
1. Then
13.5 and
This is considered to
K,D :
The unit bearing under the foundation for test conditions and one-half of maximum wind load is found to
the 1 foot, 6 inch slab for diagonal investigate "area of the lO-foot octagonal pier-is used tension, the of the equivalent square of 9'1 feet' side to compute the With t'he e/D of .159, K is found to be .88, by use of Figure 4 and. KD : 19.35 {eet. The distance from the
Abouf the Aufhor
Design and Construction, IJnion
in the United
K' :
a88)
:
68 psi
(
75 psi
:
2,900
in CC, Naval
(965
-
:6.45and K'
- (+)
:
.2e3
+
625)
:
1'310 Psf
:
120,600
Bending moment
+
J- 36,800:
(e65) : (1.08)
157,400 lbs.
(113)
10
-
(1.83) (1,310) (113)
l0
:
139,000
+
319,000:458,000 ft.Jbs.
Area of steel required per foot
458
:
(20) (14) (1.44) : 1.74 sq' in. Per ft. of width. Since wind forces contribute more than 25 percent of the moment, stresses can be increased one-third so the area becomes (.75) (1.14) : .85 sq- in. A six-inch spacing each way of No. 6 bars - .88 sq. in. I0 : 4.7 inches
:
1
lzoy
5
7.400
1rg 1.asn47;:
137 psi.
pounds.
The bending moment ;t : (6.45'z) (%) (1,310) + (%) (6.+5'2) (965; : 27,300 + 13,400 : +0,7A0 ft'-lbs'
States
Navv some of his billets were: Design'and Construction Ofiicer, Fifth
(.75) &0.7)
Druw Brown
Missile Test Center, Point Mugu, Califomia, and Public Works Ofiicer, Naval Station, San Juan, Puerto Rico, and Naval Air Station, Kaneohe Bay, Hawaii. He is a
of International Association for Structural Engineers and has BSCE and CE from West Virginia UniversitY.
26
1,475 psf.
Some foundation engineers prefer to base the steel and bond requirements on the middle one foot wide strip. Under this condition the force for bond ;t : (1,310) (6.45) + (%) (6.45) (965) :8,450 + 3,110:11,560
Carbide Chemicals Company, South Charleston. Mr. Brofn's professional experience includes Several years in the Bridge Department, State Road Commission of West Virginia. He has been a consultant on bridges for several cities. During his 12 years of
member
790 psf,
2,eoo: e65 psr
and bond stress is
Charleston, W. Va. and an engineer
Naval District, RO
:
By use of the curves on Figure 7, the shear for bond, the'bending moment and width of beam are computed for K' : .293. Width of beam: (1.82) (11) :20.0 ft. (circle): (1'96) (11) 0: 21.55 ft. (octagon) (965) (11r) r3.15) (96 Shearforbond: (.76) (1,310) (11') +
be 2,370 psf.
To
r 1.o
: (-tt-,f
",
and therefore would be a waste of time.
active duty
:
625)
/
maximum allowed.
Ps
'159.
be near enough to the allowable soil bearing' To strive for closer agreement is believed to be inconsistent with the accurac! of the established bearing value of the soils
in
+
1e.35
The section for computing bond and reinforcement is taken along the side of the equivalent square, A-A Figure
foot, 6 inches.
":#: From the curve
(790
-
\
,4"
pounds.
Next try a 22 foot,0 inch octagon with a thickness of
g.s
2,900
and P*- :fji!-^^] 2,900
By referring to the curvqs for computing shear for diagonal tension (Figure 6) and using K : .24, the width of the footing resisting shear is 1.04 (11.0) : 11.45 feet. The shearing force it : (.515) (1,475) (ll'?) + (.23) (790) (11') : 92,000 + 22,000 : 114,000
2l-9 ft.
W:
:5.28
22.0
(11.3) 3,000
3'36
:
_-
to the point where the diagonal *#* l.l7 or 5.72' Then K'D :
equipment
tension is computed
and Ps
is obtained from the curve on Figure 3 as '153. Then
1
of
s00.000
T)
center
Bridge and degrees
et : ffi
:
1'51 sq' in', afive-inch spacingeach
way of number 7 bars :
u:
11,560
1.44 sq. in.,
l0 :
6.6 inches.
:
142 psi. (6.6) (.BB) (14) The design for the top of the slab reinforcement, "top
bars," which are required by certain combinations of loading, is left for the reader. ##
Calculation Form for Foundation Design For complete design of
octagonal foundations for stacks and towers or for estimates only, this form will solve the problem easily and quickly Bernqrd
il. Shield, Celanese Chemical
Pampa, Texas
Iu rrre
Co.,
the length.
oF FouNDATroNs and. structures for chemical plants, the structural engineer normally is not too concerned with a highly theoretical, or complicated DESTGN
mathematical approach. From a practical standpoint, the ,assumptions quite often are not accurate enough to justify such an approach. Since the chemical industry is such a fast moving, often ghySinS,_ arrd complex field, the design engineelr often lacks sufficient time to make an accurate theoretical analysis or sometimes even a very thorough practical analysis. Quite often he must wade throug-h a lengthy article or text concerning an unfamiliar pioblem, Jr u design
problem which he has not worked recently. Whiie the time schedule suflers and other details of the job are neglected,
he must set up the problem for practical
analysis. For many problems of a rLpetitive nature, much time is consumed in setting up the sketches and frame_ work for an analysis rather than in simply solving the problem. many times have you heard the question, ,,When
.Ifow will the foundation
drawings be out?,, i hr.." heard it many times, quite often as soon as a request for appro_ priation for a new installation is approved. Faced with this situation, the engineer must corxtantlv seek solutions to his problems that will give safe and economical designs and use a minimum of his own time. . The following calculation form for octagonal foundations for towers and stacks was devised wiih this idea in
mind.
used the prototype of this form quite success_ . .Y"^h"": fully for about seven years ind believe it is worthwhile to pass on to others. The form is largely self_explanatory
with
the.nomenclature and design ,r,"ihod being explained
as th6 solution progresses.
Design Bosis. The following general comments should in using the form 1t firrt time. Moments are computed about the centroid "of the base of the pad, ignoring any shifting of the neutral axis as loads- are applied. Soil stresses are computed using the section the base pad around its axis of slmmetry. The T:9".1"r."j slightly- higher soil stress which would be obtained by using the section modulus around a diagonal is ignorej. Stresses causedty a moment in the base pld ur" coirputed according.to the ACI code by compuiing the moment along a line which would coincide with'the side of a square. of equivalent area to the pier. Two_way reinforce_ ment is then provided similar to the normal method of reinforcement for two-way reinforced footings. be of help
fn,computations of forces, the area and stress diagrams are divided into simpie geometrical shapes for eale in computation. The design of tensile reinforcement in the pier. is a practical rather than a theoretical approach. Anchor. bo_lt lengths hooks are designed acco.dirrg to the ACI code for _and _hooked piain bars. the length wiii depend upon the design stress used for the boltsjso if a designer wishes to use a stress which differs from that shown on the anchor bolt table, he may easily change
If he desires additional safety, he may choose to use a lower design stress for sizing the bolts #d ..r* the lengths given in the table. I prefer using higher anchor bolt
than some designers, taking adriantage of the in allowable stresses for*combinea- toaairrg /3 i, which wind is a factor. This will of course give anihor bolts which are smaller in diameter and longe-r in length. . f have a great deal of confidence in the reliability of"de_ sign stresses in steel but very little confidence in the al_ lowable bond stress for a smooth bar. Many times anchor bolts are installed without proper cleaning and with thread cutting oil all over them. So, who knows what bond stress will be developed? I believe much work remains to be done to devise. and prove by tests, a really good method of design for img" anchor bolts. In the meantime, I prefer to rlse a desi[n r.vhich I believe to be safe and economical, and ,".ogri" stresses
increase
the right of other engineers to use their own criteria. It should be noted that the use of this form is not iimited to the complete design of a foundation. Should it be desired to obtain only the size of the foundation
pad, for such things as estimating or layout, one need only proceed through.Step 5. Step 15 with Figure 3 are quite
useful to transmit information to a draflsman, and the anchor bolt tables are useful in fabrication of anchor
bolts.
Thg . .nef time you have this type of design problem, give this form a try. It is easily ,e.rised fo, .p""Ll .*"r. You may lot appreciate its merits so much if yo, orriy have one foundation to design. If you have two or more,
I think -you might begin to-like it. If you harre 50, yo,l will probably become downright fond oi itl Procedure. Considerable time and effort are usually re_ quired to make a detailed and accurate design fo. octag_ onal foundations for towers, tall reactors, preisu.e vessel-s, or stacks, particularly if the designer is'unfamiliar with the problem. Consequently, a complete design is often not made, and this may lead to either an uniafe or un_ gcgnomigal design or both. This method provides suf_ ficient design detail for a safe and economical design. A relatively inexperienced designer can use the form, but such work should always be checked by an experienced designer. This form makes such chelking easy. The
finished calculatiorr provides a neat, underst'andable, and legible record and should be maintained for record pur_
poses.
This form is intended for the complete design of foundations which have relatively large base pads in relation
27
CALCULATION FORM FOR FOUNDATION DESIGN . .
.
to pier sizes. It can also be used for foundations which have relatively large pier sizes in relation to pad sizes; however, for this condition the designer must be alert to make necessary changes in the calculations. Referring to Figure 2, the changes which will be necessary are as follows: When Dz > 0.45 Dr, d will be a negative number, and along with area (3) will drop out of consideration. Dimension C will also have to be computed by other means than as shown. Therefore the calcuiations for shear and moment as outlined is Step 7 will have to be altered. When e is zero or a negative number, the calculation {or shear is unnecessary and Step 11 may be omitted.
An
orbitrory I ft. il
DlAGONAL
TENS!ON
D6Dlc. ouleide
odd
lnrul. fL
for vopor linert lodderl
=e4(74)(%tl)
a a E
lnsul.
:
E
! a Io
in.
rl
rO
F o
DIACO}IAL
FLEXUR E
.6t -l
lrt ol EI
ta
-o F ol iF rol ro
bt) u
7/
t- 'L "ot +Cc
E F
Top
F
E!!
LEXUR E
BON D
FIGURE l-Record dimensional data on this figure.
FIGURE 2-Stresses due to flexure, bond and diagonal
ten-
sion are computed along these sections.
Step 1. Record dimensional data on Figure 1. Depth betrow grade, h5, should be deter-rnined by a reliabie soil survey for the site. The pier diameter, D2, is usually about 1'-8' larger than the tower skirt diameter. 'Ilhe pad diameter, D1, and pad thickness, h1, must be assumed and solved for by trial and error. For selecting an initial trial pad size, the foliowing method is suggested.l Where wind load is likely to govern: O, Dr : Mt : S:
:2'7{
)-
Pr:0'6
ft.
(
)-
lbs,/sq. ft.
For height zone, TB,
Ps:0'6
Trial dia.
across flats, ft. Total moment about base (See Step 2), ft.Jbs.
Structures," A58.1-1955.
lbs.,/sq.
For height zone,'12,
s
Allowable soil stress, 1bs/ft.2 (Suggest using 1,800 lbs fte for first
in the ASA bulletin to Figure 1 and Table 3 For height zone, T, Pr : 0'6 (* * Value from Table 3
,/M"
trial.)
Wind load computations are based on the A.S.A. bulletin "Minimum Design Loads in Buildings And other
28
Refer
and list wind pressures for each height zone for the plant site as follows:
(
)-
_lbs,/sq. ft.
For height zone,Tn, P.r
:
0'6
(
lbs,/sq. f t.
Compute and record on Figure 1, the values of P1, P2, La,L4 etc., for the size tower being
etc., and the values of used. Wn
: Weight of reboiler, full of water:
lbs.
Step 2. Compute total overturning moment, M1.
M'u: Moment L. (Pr) : L, (P,) :
due to wind:
ft.
lbs.
ft. ibs.
L. (Pr) L. (P,):
X
ft.
Tot.
Ibs.
M-
Mn : Moment due to reboiler weight: Wp
(C):
X
ft. lbs.
:Total:M,v*M":_
Mr
.-ft.
lbs.
Step 3. Compute vertical loads.
: Az: A,
As
:
base :0.828 (D.r) : Area of pedestal : O.B2B (Drr) : Area of
Area of
: _=_- _.' _ sq. ft. : _ .-__. sq. ft.
0.828 (-...-...-..'...-) 0.828 (..
---.-)
fill
\ : Yrr1. o1 6ur. Wr,:Wt.of pedestal :A, (hr) 150:_ Wr,,:Wt. of foundation:Wu * W. = wr : wt. of fill : A, (hr) 100:_--- x
X ._-=-X 150:_
_lbs.
* ._-:_
_=-
X
Ibs.
:
100
lbs.
Wr: Wt. of tower at time of mounting on - foundation. . . . . : We: Wt. of tower accessories and contents installed after
Ibs.
mounting:
Wn: Retroiler wt. Ww: Hydro. test water
Ibs.
Insulation
1bs.
Piping
lbs.
Platform and ladders
lbs.
Other
lbs.
lbs.
Wo, Total:
lbs,
W: Max. wt. on soil: W. * Wr + Wr + Subtract W* We : Wt. of finished tower, empty W,: Wt. of Iiquid Wo: Wt. of tower,
WA
Ibs. .--.
-
operating conditions
:
Wn
:
* W,
stress at site and
Section modulus of fdn. pad Case 1. Tower operating
:
Sz
:
: l:0.109
(D1s)
:
full and full
0.109
(_)
:
W,/Ar
ft.3
with full wind load.
Soil stress due to load
Case 2. Tower tested
lbs.
depth ,: _lbs,/ft.z
: W,/AL _/_: .: Soil stress due to wind Mr/Z : _-=--./.-=-S : Maximum stress : S, * 52 : __ /_:
S:
lbs.
--
Neglect maximum theoretical Ioading with tower wind load unless tower operates liquid full.
Sr
- lbs.
when operating
Step 4. Compute maximum soil Ioading.
Sa: Allowable soil
_-..--
Lbs./ft.2
:
lbs./tt.z lbs./!1.2
full with negligible wind.
:
IJse maximum, Case 1 or Case 2, must be
lbs./f.t.2
S
So
29
TABTE
l-fyp6
I
Bolts with t8Oo Hook See "Anchor Bolts Detailing Dimensions" on last page oI this aticle
Proiection Sta
in.
Thd.
d
Sleeve
-dic
Area
Min.
Hook A
L**
op0
in.
x
Net**
Serles
z-1.
4.9 ii ui h-E
L1A,
r%, tBA"
rt4"
tsl, 2"
L l/2" for bolts l/2" to 7/9" !t/2" for botis l" lo 2l/2"
D=6d D=8d
Step 5. Check stability. The rnost unstable period is usually just after mounting the tower on its foundation, prior to adding the weight of accessories. Erection S^io. : (W S-ro. (Soil
stress on
-
WA)/A, -S., : (lbt./fr.,
windward side) must be
-_) 2 O
This computation will give a resultant minimum soil stress which is on the safe side for stability because the vertical load is computed with the tower stripped of accessories and the overturning moment used in computing S, includes mornent due to
2%'
accessories.
Should the designer so desire, the moment can be
for this check by substituting the tower diameter for (Do+ 1) in Figure l, eliminating reboiler moment and recomputing moment on tower in a stripped condition as it normally would be during erection. If this is done, stability under operating conditions should also be checked as follows: (See Case 1
reduced
above)
S,:
-
Sz:
Subtract
]bs./ft. Ibs.,/ft.
Operating Smin.:
lb's.,/ft. and must be
Step 6. Compute dimensioru and loads for computations of stresses in foundation pad. Compute size of square pedestal with equivalent area of octagonal pedestal, A2. (See Figure 2.) IACl-318-PAR 1208 (a)]
ft-'-
a: lTr: V b:
.414
(Dr)
:
.414
d:/zDr-/ru-
ft.
+ (
ft. ft.
ft.
dr(s, g-: /r.:.t^ ) (d, is in rt. ) ,/" D, "
"
a
*
lbs/ltz
/\-
I
t_
-
ft.
+
Y:b*c L
_ o u3--_wr*wt ^ J\1
-
ft.
lb"./I*.,
-r 30
+
(d, is in rt.)
\-/
lbs/f t2
Srr:Sro-S,
:
lbs/fu2
Srr:Sr-Sro
Sn:Sr*Sr-S, I
lbs ft2
l!_,
s,,::h+#(sn)
2 d2 (d2 is in ft., here)
ibs ftz lbs/f.t2
ft.
:
.____, -
Sr:Su-Sr: se:Sn-su:
I
m
0
lbs {t2
(-)
: c:.707(b):.707(__-) :
2
lbs./ft.z
-
Lbs/ftz
TABIE
)-lyps
2 Bolrs wirh 9Oo Hook See - "Anchor
Bolts Detailine Dimensions,' on l6t page oI this aticle -
Thresds
ProJection ij1.
Sleeve
x-
Slze I Thd. Net** d I Serles Area
%'
-dio.
lzd
at
?d
4rV'o
%
_l
D=
OO ZN 4-g";
in.
t
IN
SQ. FT.
X LOAD,
(2)
:
(c)2
(2)
:
(c)'
-
(_
o,-11t
O'-L7D
o'-5,
o,-a,
0,.9,
1,.1
t,-oil
0,-5n
0'-9,
|,-0,
0.419
t,_3,
o'-6,
0'-11
1'-0'
0,.7, 0'-8"
1/t'
| .9-
o.929
181" '- II a6 q<
7'-tlr
l,-9n
o,.9,
r.155
2'-OD
1'-11
0,-70,
t,-54
0'-77'
t,-6,
I Eq
1.405 1.980
t,_0, 2'-11"
2.652
2'-9"
3,-8,
3,-6,
L'-6t
2,-4,
L'-9"
3'-10' 4'-6i
t,-4D
4.292
I
t,-5, |,-8, 2'-Ot
2,.o,
3.423
h
2'-6D
,.Building Code _ Computations are made according to Requirements For Reinforced Concrete,', ACI g15-56.
Figure
:
Ib6.
(SHEAR,
lbs.) X ARM,ft X (%c* d)
x(
X
l'rSs
X
S-
+
MOMENT, ft. lbs.
\:
x
(-.-.*_ ):
X
(t/sc
X(
I
d)
+
X (%c*
\-
d)
x (.-*_): x%d
)rx
:(dxD1)
:
X (%c*d)
XS"
X
:(dxD1)
X
,/,5,
X=--X -X
x%d
: -/\ Totals
d, req'd:
o.202 0.302
t,-5D
X '/sS*
:x
d, req'd:
o'-7,
1,-6,
)rx
B.
o,-6n
I
_ _x _.-x
Step
o,-4,
t,_6,
Xsu
:(cXb)
(3)
0,-8,
y-gn
2r4',
:(cXb)
(3)
o,-9,
--l
2, Section R-S.
(1)
0.126
0.551
aa
Mln. L*x
-t
the edge_of the equivalent square pedestal:
(1)
Y
o.728
7%" I ::'o d iiN 2D I Eir -IEJ ors --t 2%' I o@u
Sl.p Z. Compute total moment in pad along a section "See
AREA
D
].Dl
-t
a/2"
J
7%', ltz,
min.
IImk
\,/
V: .--lbs.
Mu:
ft. lbs.
Compute depth of pad required for flexure.
!
K:lc/2 kj:=_/2
/M"
KD,
J=
-V : __
k-
_._ in.
see
Fig. 1; M, is in ft.
fs
-
_-X
1+
lbs,/inz lbs,/in2 lbs.
The following factors may be obtained from tables in Concrete Design Handbooks or computed as follows:
I t-I ,';f"
f.rc:28 day compressive strength of concrete : fs : Allowable tensile stress in steel. : D, is in feet in formula,
X
: .......=--- l!s/in2 fc: 0.45 frc : 0.45 X
j:1-'/sk:ld,
f
: or ( : + 3 in. _--_
bar diameter -i- 3 in. must be
+
-.-_-
assumed
h,
in. Actual d,
used
1n.
3l
CATCULATION FORM FOR FOUNDATION DESIGN . .
.
Step 10. Compute bond stress for steel design above: 20
Step 9. Compute steel required for bottom of pad for ' flexure. (Ref. ACI 318, par t204-e)
12M.. : 0.85 .-_t2( A: 0.85_f"='-,,,^t (j) d,
)
M"
is in ft. ltrs. See Step
Number of bars across
:
p
Use
Computed bond stress, lbs./i1.2
v
: __
in. o.c.
(distribute
V
Dr. : =
X
bar perimeter'
-__._: 1bs./in.2
*Allowable p: bars at
: --_
x
D' X
>0jd,
d, is in inches. See Figure i.
Steel supplied -
:
--
7
No. -_uniformly) both ways across
Sum of bar perimeters.
: ___
in.2
-
_x_x_
:
:
lbs./in.:
Total Shear, lbs.
See Step 7.
d, is in inches, See Figure l. *Ref. ACI 318, Sec. 305
in.2 --
Step 11. Compute shear unit stress as a measure of diagonal tension along Section f-g, Figure 2. Area on
(4)
plan.
Stress indicated on stress diagram.
(5)
+
Volume of area-shess. Shear for
: s, a#
se
shape
Geometrical
(The numbers here are for defining the geometrical shapes o,n Figure 2. FilI in blanks at right for computation.)
:
(4)
-
S,,
2 s*' 2
t.l
(-+-
2
+ -
e u11
+
lbs
) ( _\
lbs
g
(-
'1
lbs
\
Ibs
\
lbs
-
s,o :t,.(?)(+)
(7)
4'-
I/_
_I_
s,, =f (1) r'r
(6)
4'-
--l
s.^
)
:
S,,
1"1,
1Cl^
(_
(-
12
shear for diagonal tension:Total of Compute diagonal tension shear unit stress, v:
m
\ /
_
12
V":Total
above------
- -)':--lbs --lbs
V",:Total shear across Section f-g (See above), m: Length of Section f-g (See FIG. 2) : ___ d" is in inches,
) (_)
Allowable v:
(__
See
FIG.
1.
)
-lbs/in.,z
Ref. ACI-318, Sec. 305.
32
:
Srr(e)2
:-l
\t
)
1"11-;
:/c-
(6)
(-
-)
:-(_2',(s)
diagonal tension
-lbs/inz
lbs inches
tn.
(150) + (100): Step 12. Compute flexure in top of pad due to uplift. W/:Figure 2, Section R-S. Wr:Uniform downward load on areas (1), (2) aod (3), See Figure 1, and using an average weight of reinforced -lbs/fu2 Figure 2. concrete : 150 lbs/cu. ft. and an average weight for : h1 (150) + hs (100) earth fill of 100 lbs/cu. ft.
AREA IN SQ. FT. (1) bXc
Xw,
XG/2*d)
x-x(2)
c2
(3)
: :
X LOAD, Ibs ft2 X ARM, ft.
MOMENT, ft.
r(-=-+-): XW/
:
X(c/3*d)
(----), x-
x(\5 ^ +-):/
DlXd
)(d/z
XW/
:
Mu, Total moment i:
Step 13. Compute steel required for flexure in top
of
pad.
:
ft.
d, is in inches. See Figure No.
IJse
Ibs.
1.
_.bars
at
--(Distribute uniformly across D. both ways)
12Mu 0-85 - - 12X fsjd, X X Mu: Moment -: due to uplift across Section R-S, Figure 1. Ars
lbs.
-0.85
--
Steel supplied
:
Number of bars
X
inches on center.
Area each bar. in.2
Step 14. Compute size of anchor bolts. Refer to Figure 1 and compute moment at base of tower as follows:
Mnw: Moment at base due to windr ft.
lbs.
ft.
lbs.
ft.
lbs.
ft.
Ibs.
M-*:
ft.
lbs.
-------:
ft.
lbs.
Mr:Tota1 moment at base:Mnw,*ME - - - - - - -: N : Number of bolts: _(,,N,, should never be less
ft.
lbs.
Mn:Moment due to reboiler from
Step 2
then B and preferably 12 or more)
Dn: Dihmeter of bolt
circle
:
ft.
Fl:Tensile force, due to Ma, per bolt:*(4 (wr/N)
:
(4X
M1,,/NDu)
-
-(-/-):_-._-lbs.
Ftr: initial tensile force due to tightening nut. - - -: (5,000 lbs. is suggested for "Fr") -/--X-.) F:Total maximum tensile force : Ft *F Fr - - - - - - -: Net Area (At root of thread) : F,/Allowable stress Size
of anchor bolt (Add /s' to
size determined above
for
lbs. lbs,
: --
corrosion.)
: Net Area of selected bolt. : See Tables
in.2
in. dia.
in.2
7 and 2 for detailing dimensions of anchor bolts, and for net areas to use in selecting bolt
xThis simplified formula is not exact, but is always on the safe
sizes.
side.
33
a o
o
i
o
()
n
long golvonized iron aleevss.
rJ'
UseJ'dao.
= No. Dio. use- J'dio. bolfs iE 3 t ies ol t2" o.c. ol *-borr -'o.c. y'-'borr olJ'o.c. Fin. Elov.
both woYs.
bolh
woYe.
EL EVAT IOT{
Step 15. Pedestal steel and reinforcing steel placement'
Thic cketch ia
for uss itl drofting finiehed drowing:'
For vertical steel in pedestal use greater of following two steel areas: (l) As: Net :
area of anchor bolts. in.z
(2) As: No. 5 bars at (max.) 6 inch
spacings.
tn.. No.
Use
formly around pedestal in octagon
As:
as
bars distributed unishown (Figure 3)
in.2
Anchor Bolts Detqiling dimensions. Tables I and 2 can be used for fabrication simply by marking or circling the desired bolt size and shaPe.
*Total
length:P*S+L+A
***Net AteA: ao, in,z ** Design basis for L: : Computations are made using allowable stress
PLAN
an
of 26,600 psi' Allow 10,000 psi for the hook
FIGURE 3-Foundation
details.
1 and develop the remainder of the bar strength in bond over length, L.
About the Author H.
ft ut fs'u' t' : rdp
is an
engineering group leader with Celanese Chemical Company, Pampa, Texas' He
Bernard
supervises
Shield
a dePartmental
grouP
hahdling mechanical design phases of plani alterations and expansions including project engineering, mechanical, electrical, and instrumentation. Mr. Shield holds a B.S. de-
gree in civil engineering from The IJrriversity of Texas. He worked with the Surface Water Branch of the USGS and instructed in the Civil Engineering Department of the Uni-
veriity of Texas before joining Celanese. He is a Registered Civil Engineer in the State of Texas, a member of Chi Epsilon, Tau Beta Pi, ASCE and TSPE.
Where:
psi.
fsrr: Allowable stress over gross area due to a":
hook
:
10,000 psi. Gross area of bolt, in.2 ACKNO\{LEDGMENT
There have been
rony fine articles published on t-his subject and I wish tc
have gained from them. In adglition, e@d @m;entr from Celanese engineers, particululy the late L'ldie {ay9{s and Willard North Carolina, 6f Chalotte. Johroon and womac soward Mth f"*.,'have contribuied to the developmqnt-o{ -thjs eim*-"i'fii[op. paper. I also wish io reognize the uork o[ our draftsroan, Don Staflord, in the drafting of this {orm,
*tr-"*I"a".- tfr. lnfo.uation which I
LITERATURE CITED
l Wilbur, W. E., "Foundations for Vertical Vessels," Pnrnoreuu
34,
34
fs: Allowable stress at root of thread : 26,600 psi. ao: Net area at root of thread, in.2 p:Allowable bond stress for 3,000 psi concrete:135
No. 6, 127 (1955).
RETTNEB'
Use Graph to size Tower Foofings Dimensionless numbers, computer
If M is increased further the structure must topple. So long as the allowable pressure is not exceeded, all of- these possible arrangementi are inherently stable. How_
calculated and plotted on graphs, simplify sizing of octagonal, square and rectangular spread footings.
J. Buchonon,
ever, at some stage in the sequence, the maximum soil pressure, th.at is the pressure af the extreme point on the leeward side, becomes equal to the allowable pressure. As the moment is increased further, the pressure at this point exceeds the allowable and the structure is in danger of toppling caused by differential settlement.
Newcastle University College,
Newcastle, N.S.W., Australia
Gnaprrs oF srMpLE dimensionless numbers may be used to size spread footings. These numbers describe the action ot the footing under a known Ioad system and allow the user to select a footing size that wiil maintain stability without exceeding a specified maximum allowable soil
For any particular footing shape (square, octagonal, etc.) and. orie.ntation, a pressure pattern as shown in urry o1e .of the diagrams_ of Figure 2 prescribes a unique relation bgtwegn W, M, the maximum pressure p, ,La the plan size of the footing-described by some character_ istic dimension L. any such case the dimensionless -beF-or formed from these
Deanng pressure.
. A typical footing arrangement is shown in Figure As in the usual
wLWsW
treatmeni the soil under tfre fo&;ng is When the moment
(M)
-M-, rrp, ]f,i_, etc._
1.
be perfectly elastic, and no credit is allowed for .1".
llk"" tne sorl tateraL support.
groups, which may
variables-
is negli_
have fixed values.
The graphs-of Figures Z and. 4 show the relationship ,between two of these groups,
w
/w
_w
r/ " and --_Pt.2 for footings having square and M
',
octagonal plan shapes and tne onentatrons shown. The terms of the first g.roup and the footing shape are the design data. Calculution tf this group and reflrence to the graph for the footing ,t apJ specined gives the value of the second group fiom it i"t tfr" size of the -
footing may be calculated. _ Figure 3 includes all the cases where the whole of the base of footing is loaded as in Figure 2(b) ; that is, -the where there is some pressure over the i"hol" of 'tLe lower face of the footing. The upper limit is at the point where the minimum pressure is 05 percent of the maximum.
B?9"9 this,point the effect of th" moment load may safely be neglected. Figure 4 describes
FIGURE l:Typical footing anangement for tall
towers.
gible compared with t_he dead weight (W), the soil bear_ ing pressure is uniform over the *f,ol" u.# of the base of the footing as shown in Figure 2(a). As the moment
increases, pressure diitribution'"h*.rg"r, as shown .r.iJ in Figure l!: 2(b), (c) and (d), until it reachei the extreme impossible)'case shown in Figure 2(e) 1111-r:,rr:ctice where the structure is just balanced on one corner of the footing and the bearing pressure is infinite at that point.
the cases where only part of the base is Ioaded fas in Figure 2(d) ]. The lower limits cor_ respond with cases where only about one tenth of the
base area is under load. Actual designs will rarely ap_ proach this condition or go beyond it. fne ,pp", iiroii, of Iigure 4 correspond, o1 cor.se, with the lower limits
of Figure
3.
given loading system on a footing, there is, except ? . Fo. Ior a ctrcular tooting some critical orientation of the axis of rotation which produces the highest maximum soil pressure. For both the square and the octason this orien_ tation is the axis passing through two verlices. The curves for the octagonal footing have been calcu-
35
(d)
(c)
(b)
(a)
FIGURE 2-Changes in soil pressure for tall towers'
lated on this basis. Similarly, for
tions, values of the parameters zl and p have been added to the graphs, where in Figure 3,
a completely unre-
strained structure on a square footing, the curves for the afi.""f axis should be used for calculating the minimum size of footing. If the struiture is restrained so that rotation about only orr" t*i, is possible, as for instance in the case of a pipe rack standird, a more economical design results if a to ;q;;t; footing is arranged so that this axis is p-arallel calcufor used one side. ThJ approprilte c,,rve is then i;;i.;;f the fooii"g tir". ln this orientation the required
minimum
and in Figure 4,
!:
to the axis of rotation' its - fieater'side perpendiculai well for design of recequally used b" fir" curves *uy q the plan aspect factor the using by footings tangular ratio of the footing. oi rotation ,* - dimension perpendiculal t9 the.axis diinension Parallel to the axls
In this
these cases,
L: dimension perpendicular to the axis of rotation Usually a will be greater than 1, but if for some other reason a'rectangular-footing must be laid out so that a is less than 1, tfie graphs may be used in the same way' The procedure then is to find the size of a square footing *hi"h, with the same loading, would produce a maximum bearing Pressure
"f I
matically by using the grouPs,
frtit is done auto-
M lwt
w{
-F-
and
Wa
5p
As is usual with dimensionless correlations, the units of the terms must be consistent. The length unit will usually be feet; the force unit may be pounds, kips, tons, or any other convenient' Typical sets are:
lb. kips tons
ft. ft. kiPs ft. ton
lb'/tt''
Ib.
kiPs/It'z
*o/t"
When the size of the footing base has been calculated
it is necessary to calculate the thickness and reinforcement necessary to resist the shears and bending momelts
in the footing itself. For
w /E *r, -'fr'YE
^ *,i**
is les thm
convenience
0.73, the situation
in
these calcula-
is
revened,
but in
36
connection
it
should be particularly noted that
*h"; " portion of the footing is unsupported by soil-re;;i;" tiere are shears and, more important, bending moments in the unsupported section of the slab in the oplotit" sense to those-usually considered-in the design' t'Iiere st."rses must be evaluated and the slab design may reouire modification to resist them (e'g' by the addition of iop bars to resist the reverse moment) ' In common with all other methods proposed for esti*uii"g footing size, the calculation must be a trial and !ro."tt.
The known data are usually:
"rro, . Structure deadweight ("mPty, working and under hydrostatic test conditions,
if required)
;
o Pedestal size and weight; o
this a
Depth of footing base below C."Pq (from knowledge of trost Iine leve-I or situation of desirable load bearing strata) ;
o Allowable maximum soil bearing pressure
(P)'
For full details of estimation of these see Brownell and Young2 or Marshall.s The moment load (M) 9a1 t{en be caTculated from the wind load and depth of footing base below ground. The total deadweight (W), however, comprises, i'esides the weight of the structure and the pedestal: o The weight of the footing itself, and
The weight of overburden above the footing' These can only be calculated when the footing plan size and thickness have been fixed. Thus it is necessary o
rectangulrfmting with uis of rotatiou parallcl to *"i side shoulil still be red.
ProPortion of the width of the
o Wind and other eccentric loads;
as
marked on the graPhs'
LWMP ft. ft. fr
ttne
footing under load'
Knowing the value of the appropriate p.arameter and of th" *rii*r* soil bearing pi"tt*", P, the load distribution over the lower face o1 the footing and the rethen be calcuili;;A thickness and reinforcement ormay Brown'l Marshall3 of methods Utea Uy the
size of the footing is somewhat less'* --irr r,rch a situition, however, an even more economical d*ig" *ry ,"rrlt from using a rectangular footing with
In
Pressure
'-;;i;;;;;;G
to guess initially a footing size so that (W) can be estimated, and then to refine this estimate by trial and error. Often, and particularly for deeply based footings, the siab thickness is of minor importance at this stage, since extra thickness of concrete onlv displaces overburden of not greatly different density. If the initial estimate of the thickness is reasonably good. final adjustment will have no great effect on the deadweight (W). It is usually desirable to compute separ.ately the footing _ size required for several critical load conditions. These are:
. Minimum weight and maximruu nind effect, e.g., in course of construction;
=l
;
o Working weight and maximum u'ind; o Test conditions-filied with tater ancl 50 percent of maximum wind moment.
The evaluation of the first and last of these depends on the design and method of construction, and no useful general rules can be given. A reduction in wind load for test conditions is allowed since it is most r.rnlikely that the test period and the maximum rvind would coincide. Since tho construction period is normall1, much longer than the 456
8t0
20
30
40
lwlE
60 80 t00
200 300 400
ur'p
FIGURE S-Relationship between two dim,ensionless numbers for all cases where some- pressure acts on lower base face. 0.5
0.5 o.4 0.3
l'
I
0.2
*l-'
o.rs
P
W: Weight of ,structure, footing and overburden M: Moment of wind load and any other eccentric loads about the center line ol the base of the
0. I0
0,09
'0.08 0.07 0.06 0.05
0.04
0.5
0.6
0.8 I
P: Maximum allorvable soil bearing pressure L: Characteristic length: square-length of side octagon-width
across flats
rectangle-length perpendicular to the axis of rotation (See Figure 5) PIan aspect tatio of rectangle
_ Length perpendicular to
axis of rotation
Length parallel with axis of rotation (See Figure 5 )
For other cases, a: 1 Minimum Dressure ,z (.tlgure Jl :: - _:+- pressure lvlaxlmum p (Figure 4) : Plgp-crrJion of the rvidth of the footing which is under load.
1.5 2 w
footing
3
4 5 6 78
/TE-
M/P
FIGURE 4-Relationship between two dimensionless numbers for cgses pressureicts on only part of Iower base face. ^w-here (i.e. Fie. 2d).
test, no such allowance is possible for min. wt. and max.
rvind effect.
The procedure to be followed is iliustrated in the fol-
lowing examples.
Exomple-Ocfogonol Tower Footing. A footing is to be designed to carry a. tower 54 feet high and 4 feet in diameter to be placed on soil for which the maximum allowable bearing pressure is 2,000 lb/ft., The frost line is 4 feet below grade and the pedestal
top is to be 1 foot above grade. The footing
base 1s
made
37
SIZING TOWER FOOTINGS. .
5 feet beiow grade, i.e., 1 foot beloru the frost line' The design" maximum wind velocity is 100 mph' The ma*"imum wind moment ahout the base of the footing is calculated to be 200,000 ft' lb' (M)'3
Restrained
=
0.318
L : 8.86 ft. so that rotation
78.6 It,2
abor'rt onlv one axis
\V PL:
30'000 lb' Empty Tower 9,000 lb' contents working and Appurtenances 40,000 lb' Water fitt for hydrostatic test
For a pad estimated to be 13'5 feet across flats and 1 foot thici< rvith an octagonal pedestal 6 feet across flats + i.", deep and ctay nit of density 90 1b'/ft'3, esti-
)\ -"
Lr -
0.38t)
T, :
8.1
. 65.8
ft.:
ft.
u,eights are: 63,000 lb. 33,000 lb.
Concrete Fi11
Then calculations for the three critical conditions are:
EmPtY Working Test 126,000 13s,000 i66,000 w (1b.) 200,000 200,000 100,000 M (lb. ft.) 2,000 2,000 2,000 P (lb./1t.2) \v i* 0.63 v-63 u.G.s v-G,s 1.66 r/ &3 M\ P : 5.54 : 15.11 :5'00 W ,f..r, nraoh) 0'405 0.585 0.423 PL2 ' 83 67.5 63 L2 U.5u5 0.423 0.105 11.9 12.61 t2.47 L (ft.) Thus the assumed size is too large and could be reduced. The next trial rvould assume a l2-foot octagon' and the minimum size would probably be somewhere
t
FIGURE 5-For rectangular footings, "a" is usually
greater
than one.
as in Exomple-Rectongutor Fooling. For the loads : (short 3 a assllme above, e*ample footing ,h" ,qrur" side parallel rvith axis):
]t - /'E M\ P
,o.i rl x
wu
:
PLT
:
L.
Unrestrained. Diagonal axis
25
:4.34
0.505 (from graphs)
7.5.
=:
1'18.5 ft2
0.505
L :
12.2
1:
+.1ft'
ft.
The rectangle required is l2'2 ft' x 4'1 ft' having pian +g.S ft"a, ,gri.rst 65.8 ft2 for the square footing
(from graPh)
"r"" under the same conditions.
In each of the above examples the maximum pressure *itt U" equal to the allowable, and the pressure distribution may te immediatelv sketched after finding the value of the parameter p o, ,i fronr the appropriate- graph'
Aboul the Aurhor lecturer in
Considering the essentially rare -and transitory occurrence of the riaximum moment load, the- basic assumption t"ginnitg are sufficient for most applica;;;-;i;he iio".. ffr" ,rsrr..r!tio, of perfectly elastic soil, however' is not entirelv sonlnd and in tritical cases the advice of a soil mechanics'expert should be sought'
chemicai engineering design at Newcastle l]niversitv College of the Universitv of New South Wales, Tighe's
Hil1, N.S.W', Australia' tle holds B.E. (Chem.) and M.E. (Chem') degrees from the University of Sydney'
Buchanan held Positions as a design engineer with I\'[onsanto Chemicals Ltd. and Union Carbide Ltd. in SydneY Prior to accePting
Mr.
his present Position'
I
o
* -/* :0.5t/-Ii:2.5 M!F
a
cl
t
bearing pressure of 2,000 Lb.lft2.
John Buchanan is
L
o o
Exomple-Squore Tower Fooling. A square footing is to be designed to carry a total estimated deadweight of 50,000 po.,nds and a maximum overturning moment of tOb,OOO Ib. ft. on soil having a maximum allorvable
#:0.318
-T
C
.9
near this figure.
38
is
possible.
Tower weights are as follorvs:
""J mated
?5
,-*
.
LITERATURE CITED 1
Brou'n, A.
141 (1963).
Buchanan
A'
HvorocangoN Psocrssrro &
Plrnorruu RsrrNet 42' No' 3'
",'d,.*"'"ir. L. E. and Young, E' H' in ^-"Process Equipment Design" )' ch#ri" ti, ^u"i'v*1. l"t " Wile"v' and Sons37;( 1959 No' 5 Desisn Suppl' (1958)' sMarshall. V. o. l'nrnorru*huu'"""
Simplified Design Method for lnfricate Concrete Column Loading Combined biaxial bending and axial Ioad on reinforced concrete columns present difficult design solutions. This method bypasses the usual tedious computations E. Czerniqk, The Fluor Corp., Ltd., Los Angeles
IIr,RB's A NElv AND srlvrplrrrED METr{oD of solving concrete column problems consisting of an axial load combined with diagonal bending. The method can be used to determine the combined stresses and the eccentric-load capacities of reinforced concrete columns from known or assumed positions of the neutral axis. The approach is unique because it provides greater accuracy with less computation than methods used up to now. It considerably simplifies the stress analysis of many structural components used in Hydrocarbon Processing Plants, e.g., pipe supports and rigid frame structures for supporting exchangers, compressors, etc. The method bypasses the usual, time consuming, tedious computations of principal axes as well as the need to rotate all computed properties about the principal axes. Significantly, the method is valid for both elastic and plastic stress distributions. ft thus unifies in one, simple approach the straightline and the ultimate-strength methods now used in reinforced concrete design. The methods of analytic geometry and the basic equilibrium equations from statics may be applied to a.variety of problems involving stress analysis. The term 'analytic, preceding'geometry' implies an analytical method, wherein all results are obtained algebraically, with any diagrams and figures serving merely as an aid in visualiz-
ing the problem. All given data must, therefore, be exin coordinates with respect to a suitable set of axes (preferably selected so as to make the coordinates as simple as possible). The procedure will be illustrated by the rather intricate problem of axial load combined with diagonal bending. In general, when bending in a concrete column occurs pressed
about both coordinate axes, and there is tension on part of the section, the effective portion of the reinforced concrete section (transformed area) resisting the applied load is not symmetrical about any axis. Though the unit stresses may still be expressed by the well known formula: P _.M*"*oMr",
Ar*I,
such a process is rather laborious because all the values
must be related to the principal axes through the centroid of the acting section. Thus, for each assumed neutral axis, one must repeat the numerous and tedious computations of : the centroid of the acting section; the orientation of the principal axes; moments of inertia about the principal axes; and, not the least, the calculation of load
eccentricities with respect to same principal axes. No wonder, then, that 'exact' solutions have been consistently avoided by practicing engineers. The technical literature, though abundant in advice on the 'how to' side of problem solving, is extremely meager when it comes to specific examples, except maybe for the most simple cases.
Should You Trusf Compulers? The increasing use of digital computers has somewhat improved the situation. Computer programs are now available that can accomplish the tiresome solution through successive approximations, at extremely rapid rates. Ilowever, when ihe engineer views the computer output sheet, he may sometimes bewilderingly wonder just how accurate these results really are and whether he could and should put his trust in the modern marvel of technical automalion 'design via computerization.' Needless to say, the engineer has no right, nor authority, to abdicate his responsibility for professional judgment. The responsibility for structural adequacy must always be his, irrespective of the methods or tools used to come up with the answer, be it a slide rule, desk calculator or a giant electronic computer. Ifence, if he is to make the most out of the new to-ol, he must possess some simple means for spot checking the machine. In the case of biaxial bending on concrete columns, the method outlined below could probably serve such a pu{pose.
Two Design Methods. The Building Code requirements
for reinforced concrete (ACI
31S.5G)
permiti columns
subjected to combined bending and axial load to be in-
vestigated by two methods:
o The so-called elastic method in which the straight line theory of flexure is used, except in regard to compressive reinforcement. o The ultimate strength method on the basis action.
of inelastic
Ultimate strength design is relatively new in American Codes, and hence some of the old timers may feel ill
at
ease 'r,r,ith new concepts and new
criteria.
It will be
shown, through illustrative examples, that the same approach applies throughout the full range, from elastic
to elastic-plastic and ultimate strength considerations. In the straight line stress distributionmethod. the code requires that colurfins in which the load p has analysis
an eccentricity greater than
/3 the column
depth
t
in
39
]NTRICATE CONCRETE COLUMN TOADING
...
,
either direction, the analysis should be based on the use of the theory for cracked sections, e'g', that the concrete does not resist tension. This e/t allowance does not apply in ultimate strength design. At ultimate loads, flexural tension
that
it
concret-e is insignificant, and the Code requires be comPletelY neglected.
in
filethod of Anolysis for Rectongulqr Sections' In the one case of rectangular sections, it is convenient to choose two with coincide axes corner as thelrigin and Iet the sides of the rectaigle' In Figure 1: O, B, C, and D are of the liven conirete section' Line QR desig,h" "o*"r, the neutral a*is 1line of zero strain), and intersects nates the x and y axes at a and b respectively' Let the toordinates of the eccentrically applied load, P, be i and !; and the coordinates of any given reinforcing bar, of area Ai be x1 and Yi' The" inte.cept form of the equation of the neutral axis QR is:
i**:' fi
Now, assuming that the stress at any point (xi,yi); is proportional ti its distance from the neutral axis, then Uy'*,it,iptying fo, the stress at origin (o, 9), by the ratio oi tf," dittr"Jes of point (x1y1) to that of (o,o) we obtain the general stress formula:
r,:t"\ (t -+a - ItJ b/ The engineer need not keep track of. the sign, as the stress foniula will automatically result in positive stress, or compression, for all points lying to the left of the neutral axis (see Figure 1) and negative stress, or tensron beyond the neutral axis' The coordinates of the centroid of the triangular area under compression, OQR, ate a/3 and b-/3' Hence, the average compression stress within the effective concrete section
will
simPly be:
x: a/3
l-In
origin.
In the general .case, when line concrete section (see Figure the outside partially QR'ir must be subtracted from triangles smaller more o.r"'o, fr, the over-all larger one. This is illustrated in the following
compression, is a triangle.
examples.
Reinforcing Steel Slress. The stress in the reinforcing bars is obtiined by multiplying the value (f1) in the seneral stress formula, by the modular ratio, n, the ratio It tt modulus of elasticity of steel to that of concrete' " 601 of the ACI Code gives the ratio, n, as equal Section
to 30,000/f'". H".rc",'thl
:
(Average Stress)
X
(Area)
:'/s oX /z ab: t" + f
rrci
axes are: fo
ab
(in x direction, about v
-e- X t:r.* f
(in Y direction, about x
axis)
axis)
(The reader may note that a2bf24 and abz/24 are simply ihe ,ralues of the section moduli of the effective concrete area, in the x and y directions respectively') InFigure 1, the intercepts of the neu-tral axis, line QR, are shoi., smaller than the corresponding dimensions of the section. Hence, the effective concrete area under
40
----t rrri
: '^'.\' nt^( t- f.:_ - +) b/ "
-
Ai
is:
F"r:f.rXA,
,irr"'A'
N : total number of bars.
Similarly the steel load moments about the coordinate axes
will
be:
M'"":
N
i ,",*, and M'.r: i=1i t-f
u",r,
The above formulas are completely general and the
he so wishes, may use diflerent diameters for thJ individual bars, and he may or may not arrange the bars with symmetry about either axis. However, since the same modular ratio n was apptied to both tension aird compression bars, we did piesuppose -that the bond between the steel and concrete remains intact, and they deform together under stress. That is, the steel in the comp.essio-n zone can withstand a stress only n times that in the concrete. In reality, this is not exactly so' Because engineer,
M'"r: ,+
any bar A1 designated by coordi-
and the load in said bar, having an area
where
can also be shown that, for equilibrium, the load F" is located at coordinates, af 4,bf 4;hence, the moments oi the compression load in the concrete, about the x and M'"*:
in
F":
It
y
stress is:
nates xi and yi
and the total compression load in the concrete: F"
rectangular sections, choose one corner as the
The total load carried by the reinforcing bars (tension and compression) is the summation of the loads in the individual bars:
f,,:fo O-%-/t):/s1"
v:b/3
FIGURE
if
I-
FIGURE 2-Example
about centerline.
l.
#8 Lorr
Find eccentric load p and moments
of plastic flow in the concrete, the compression bars are stressed more than indicated by elastic analysis. Codcs have recognized it, by assigning higher values to the reinforcing bars in the compression zone. Section 706(b) of the ACI Code requires that: "To approximate the effect of creep, the stress in compression reinforcement resisting bending may be taken at twice the value indicated by using the straight-line relation between stress and strain." Ilowever, in permitting this use of 2n the Code limits the stres.s in the compressive reinforcing to be equal to or less than the allowable stress in tension.
In the examples that follow, the stress in pressive reinforcement shall be made equal to: f",
:2nf, < f,
(Compressive reinforcement oniy)
F"r:Ai (fri-fr):A,
X , Y
a (c)
concrete and steel reach simultaneously maximum allow_ able values (balanced design). To determine the value of fo, compare:
a
l..-\
1+-'
__
hence, use: fo
b
:0.45 f'"
fi .45f'nn
(-f, ). Ffence, use:
ft
.
\
(+*+-,)
Exomple I. Find the maximum value of an eccentrically applied load P, and the moments about the centerlines of the column shown in Figure 2, when the neutral axis is the position indicated. f'" : 3,000 psi f, : 20,000 .in
ps1.
n
:
30.000
fo
<
0.45
,1_: X
3,000
ft =
20,000 Psi
a:
12"
Ai
10
<
:
f'"
"n
0.79 sq.
inlbar
X t5 :
20,,
1,350 psi
(given) O:f#,
TABLE I--Coordinates, Stresses, Bar Loads and Moments
for Example
1
M'",
t (In-kips)
I"i (psi) 0 0 2. I7 I 7,
4
0..15
:
hence,
2
hence, use: fo
ft
.45f"n
_f
1
(b) Allowable concrete and steel stresses are reached simultaneously when:
r
'r , 'o(r:-+)n a b)
0
f .45f
f
tension bar,
Concrete stress governs when:
X,Y-
I
When this happens limit the stress in the extreme tension
bar to
B
ab -
-,
Solution:
Governing gt1s55-(6ncrete or Steel Tensile. In the general stress formula for fi, the stress at origin (o, o) is designated as fo, and is the maximum compressive stress in the concrete. According to the ACI Code, Section 1109(d): "The maximum combined compressive stress in the concrete shall not exceed 0.45f.,. Foi such cases the tensile steel stress shall also be investigated.,, Ifence, fo will equal to 0.45f"' only if concrete governs, or when
farthest away from the neutral axis.
_
i-+a>r-1-
(2n- 1) fi
("/al y/b) to 1 | (f ,/0.45f,.n) where x and y are the coordinates of the
b
concrete
Tensile stress in steel governs when:
the com-
where f1 is the allowable tensile unit stress in column reinforcement. Also, a correction shall be made for the concrete area displaced by the steel bar by subtractins (fr) from the steel stress in said bar. Hence, comp;ssive load carried in bar A; will be:
(a)
FIGURE 3-Generally QR is partially outside the
section-
+1350 +337.5
+ 4,500 -1462.5 -14,625 7,875 -+9oo787.5 -+18,000
+
+225
Load on reirforcile bars Load on concrete e?ective section
8.45
-202.13 -108.85 + 33.78 0.88
-+53.16
-268.75 +161.37
t
(In-kird
+
42.%
-t44.38 tD.m
-+
33.7E 83;90
-+256.35
Total * Loads in compressive reinforcement corrected for area of concrete displaced by bar.
4t
INTRICATE CONCRETE COLUMN LOADING
.
after bending, we write the general equation of the in intercept form: xvz I abc -t--r-where a, b and c are the intercepts of the plane on the x, y and z axes, respectively. It is immediately apparent that z is a measure of the strain, and the constant c is the maximum strain, which conforming to usual notation, may be written as eo. Constants a and b designate the neutral axis as before. Hence, the general relationship for the strain ti at any point x1, yi may be written
strained plane,
r,:r"
(r
-#-+J
Coordinates, stresses, bar loads and moments are tabulated in Table 1. Also see Figure 3.
The load and moments in the concrete are calculated next.
1.35x12x20_ 0.34x3x5 o , _ "
-
:53.16kips
6
:-E-x 0.75 : 161'37 in-kips 54.0 x 3.0 - 0.84 M'"r: 54.0 x 5.0 -- 0'84 x 16'25 : 256'35 in-kips
as:
M'"*
8i:
€o
(,_+_+)
Multiplying both sides of the strain equation by E" we obtain:
v:
+
-J-qrq"
52.28
The eccentricities of the load with respect to the centerlines of the concrete section are:
E*:
+ 2.05 : 12.05" -3-29:4.27"
10'00
Ey:7'50 Results:
P :52.28
M*: My:
52.28
,,:r"(,-+-+)
172.45
kips x 12.05
52.28 x4.21
:
630 "k
:220 "k
It should be noted that in above example the ratio e/t is less than 2/3 in either direction, and according to Section 1i09 of the ACI Code could have been analyzed as an uncracked section. The example was selected on purpose, so that the interested engineer may comPute, for the g.os, transformed section, the value of the maximum allowable load at the same eccentricities, and compare it with the 52.28 kips calculated for the assumed cracked section in Example 1. Uttimole Strength. The term "ultimate strength design" in reinforced concrete denotes an analysis based on inelastic action. It focuses attention on ultimate rather than design loads. As in elastic analysis, it is assumed that plane sections normal to the axis remain plane after bending, and as is common in a reinforced concrete column, tensile strength in concrete is negiected. The departure is, that stresses and strains are not proportional
at ultimate capacities. Section (A603) of ACI
Code
permits "the diagram of compressive concrete stress distribution to be assumed a rectangle, trapezoid, patabola, or any shape which results in ultimate strength in reasonable agreement with comprehensive tests." Furthermore, it iimits maximum concrete strain eo to .003, and maximum fiber stress in concrete to 0.85f"'. The stress in tensile and compressive reinforcement at ultimate load is limited to the yield point or 60,000 psi, whichever is
The engineer should not have any qualms about using the constant E" at ultimate strains. Since the stress in the concrete shall be limited to 0.85f"', any hypothetical stress above this value will be subtracted' The equation of the line for which f1 reaches the value 0.85f'" may be written by making fo equal to eo X 1000f"' (Note: E" is assumed equal to 1000f'"). or
f'.
0.85
:
eo
X
1000 f'c
which for the specific
as before.
From the assumption that plane sections remain plane
42
case
of
eo
:
,\
(Note that
j-+lo:r
f'"
cancels out)
.003 reduces to:
from which the intercepts on the x and y axis are be
X, :
Stress
0.717a and
Y, :
seen to
0.717b respectively.
in reinforcing bars: r",
:
nro
(, -f
-;)
= r,
About lhe Aulhor Eli Czerniak is a principal design engineer with The coordinates computer Anseles. -Lle Angeles. He coordrnates uor Corp., Coro.. Los Ange Fluor applications for the Design Engi.""rino T)ent.- reviews manual manual techniering Dept., m( methods new lrltrLrruus elops rlcw develops niques ano nlques and oeveroljs
procedures better adaptable to and pr
systems conversion
in
automating
the design and drafting of refilerY 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
smaller.
Now, when the position of the neutral axis is known or assumed, the magnitude of the ultimate load P, and its eccentricities, which result in the prescribed limit strain, may be easily determined by using the same approach
/ " \, -;-;-/
technical articles.
He has
field experience as a civii
had engineer
and woiked in design and drafting with Arthur G. McKee Co. in Union, N. J., for two years before joining Fluor in 1953 as a struitural designer. He soon headed up the structural design and drifting on various projects until assuming
Czerniak -
his present position.
\x.
/B
,/-'t', FIGURE 4-This drawing helps visualize the problem in Example and, as before, a correction for the concrete area displaced by compressive reinforcement shall be mad.e by subtracting the concrete stress from the steel stress, when determining the load
2.
Exomple
in the compressive
Tabulations of the calculations are given
*:-lgg-:-1'oB" - 215
bar.
Compute the Ultimate Load
p, and its
+
pii. Figure 4
fo
l
:
.003
TABTE
x
1000
x 3000:9000 psi : 9.0 ksi
Pointl
2,550 psi
fo
0.85
f'"
:
9000
:
2,550 6,450 psi of excess stress)
-(maximum value-
_9.0x12x20 2.25x3x5 o 'rrc-----l---6 -
:
xi I
vi
0 I0l0 B l0l1s I 8.601 o o 1 Jt 2.5t114.3 12. 2 I rz.sl 12. 3 I 17.51 2. 4 I 2.il 2. Xu Yu
xi t---12
Total
6.45 x 8.60
x
1.1.34
E*:10+ 1.08:
ll.0B,,
Ey:7.5
,
-4.86:l.gg,,
+1.00 +0.25 +0.283 +0.283 +0.167
-1.083 -0.583 +0.667
Example
Ultimate Stresses
yi
fi
psi
20
Concrete psi
Steel
2550 2250 2550 2550
+150(
1500
+15,00(
::::
-40,00( -40,00( +40,00{
+900(
1600(
2550
M"y
F,r kips
psi
+225( +255( +255(
Load and moments on steel Loads and uoments on cotrcrete
I-oad on concrete effective section
Table
2-Loads on Steel and Concrete for
Yu:0.717 x20:14.34"
/_i!!in triangle OXuYy concrere stress equals 0.85r"
in
is
Solution:
X,:0.717x12:8.60,,
967
Y:--199--+4.86"
eccentricities with respect to the centerlines of the sec_ tion, for the neutral axis given in Example 1. Use yield
point of reinforcement, f, : 40,000 drawn to help visualize the problem.
2.
*
553 553
+ 134 - 395 -79 +74
+221.8
-1005 + 790
266 -+1233
+198.9
_
-*
_
ro.z, 31.6 31.6
zs.o, 22.9
-
2t5
+
967
* Loads iu compressive reinforeement corrected for area of concrete displaced by bar.
:221.80k and the moments about the coordinate axis M'"*
:
M'",:
360x 360 x 5
3*
-
5.63 x 0.75 5.63
x
16.25
-
132.57
-122.57
x2.lS x
:
3.58:
790',k
1233,,r
Hence, for the concrete section shown, a maximum ultimate load of 199 kips (divided by the proper load factor) may be placed at distances 11.0g inches and,2.64 inches from the centerline. !JL #1+
43
;,St,,
$r
U#idAI ,*idffion
desig lii':TALt ..
T
Close centerline distance, high towers, weak clay soil and hurricane winds gave Phillips some interesting probtrems Edword V. French Phillips Petroleum Compony, Bortlesville, Oklo
SEVERAL UNUSUAL conditions faced Phillips' engineers in the design of a common foundation for two iall fractionating towers. The towers wcre to be located in the Phillips' rcfinerv ncar Sweenv. Texas.
The design conditions presented these diffir:trlt problems:
. The tou,ers
\\'ere fairly high and clos': to one an-
other. o The soil consisted of a relativcll' u'cak clay. o Horizontal forces werc to be based winds and aerodynamic vibrations'
on
hurricane
!:l
o Each tou'er had to be structurallv inclependent of the other and each self srrpporting. ,.'.:
,, Loyout Sfudy. A study of proposed lavouts indicated that it was economicalll' advantageous frorn a piping . ', vier,r,point, to spacc the torvers close to ont: another'. An s inve;tigation shorved that for indepenclent Iottnclations. 'll
ffi
ffi
'{R€.,
!.jiT*?.,1 f::il.q*ttl"l
FIGI]RE
44
l-Gin
pole supports tou'er as
it
is raised free of qround'
". '*";* .
' d.* *
#";-t:
4 7t'ffi..2:.. .a^rui
..n o Verticol
*.i;1i
:::Ytfai.=:-:1,l1r,i
-'
keyed f # iii#3 *%.i€ ^ a ^W 'u'G* *, joint construction tie mot #
***
.0
bors qnd
llild*ry*" *&g-ond Jffi@-.,
"*%Y#
.* **ffi efi.W
pedestol rogerhe,
FIGURE 2-This is the foundation after the fust pour.
octagonal mats at least 40 feet in diameter would be reqrril6d and that any spacing of about 40 feet or less would involve a combined foundation. Although this investigation indicated that there would be no appreciable economy in materials using a com-
bined foundation, one advantage was apparent although somewhat unmeasurable. It is possible that the toweis may vibrate when subjected to steady winds of 35 to 55 mph velocity. A natural vibration period of 1.0 second per cycle was calculated for the shorter tower and 1.4 second per cycle for the taller one. Assuming aerodynamic vibrations to occur at these frequencies, im_
pulses transmitted by either tower into a corfinon foundation would tend to be damped by the effect of the unlike opposite tower. This -damping would be effective to some degree whether or" o, both towers were in motion or regardiess of wind direction. In view of these factors, it was planned to space the towers on 28 foot centers using a common foundation. This spacing allowed adequati clearance for erection arrd maintenance operations. A pian and. elevation view showing the arrangement and general details is shown in Figure 3. Soi! Conditions. The soil at the foundation site is cohe_
sive. Borings were made and laboratory tests run of the soil samples. A 5-foot top stratum corriirt, of black and tan organic clay. This is underlain with 2 feet of stiff
tan inorganic clay beiow which lies clayey sand and sand.
Using shear strengths indicated by the tests and Terzaghi's bearing capacity equation, the allowable soil pressure of 4,000 psf was determined. This was based on a safety factor of 2 at a 7-foot depth with no increase permitted when combining wind and vertical loads. After the mat had been sized, uniform soil pressure due to vertical loads totaled only 1,500 psf. domputed total settlement was consequently small ind a major percent_ age of it could be expected to occur during construction. Lood Combinqfions. The effects of three separate combinations of vertical and wind loads from the towers were investigated:
I. Vessels ready for operation plus full wind forces but without operating liquids. 2.
Vessels operating plus
3.
Vessels ready
full wind
forces.
for operation under water test con-
ditions without wind.
Tower Fqbricqtion. Schedules controlling tower fabrication and tray delivery were coordinated so that the trays could be shop-installed. Also, platforms, Iadders and most piping were scheduled for installation immediately after the towers were to be erected. In addition, backfill material was to be placed before the towers were
45
erected. Since the likelihood was remote that both towers would be left stripped down for an appreciable time, no "erection" condition was considered other than to check for stability. Wind forces were computed on the basis of 125 mph maximum gust velocity at a 30 foot height. Height factors were then applied which gave the following pressures in three height zones:
0- 50
feet
. 100 Above feet
51-100 feet .
.... 52 psf .. . 62 psf . 72.2pst
-T-
= E
o b
g*=
s;f
-o ti-
nl t
These values represent pressure against flat surfaces. shape factor of 0.6 was applied to compute the pressure against projected areas of cylindrical surfaces.
A
Anchor Bolts.
In
designing anchor bolts, the upper
pedestals were analyzed as cantilevered flexural mem-
with combined bending and axial forces. in the concrete and tensile stress in the anchor bolts was then calculated according to the theory of flexure for concrete. Both carbon and alloy
bers, loaded
Compressive stress
for bolt material, but investigation showed that an alloy steel with higher allowable stresses and less tendency to creep under load was the most desirable. The practical limit on the number of bolts that could be placed around either tower perimeter was aPproximately 36. This in effect established the total tension force which each bolt must resist. Maximum bolt diameter was not a limiting factor, but by using the higher allowable stress of alloy steel a substantially smaller bolt could be used. This was advantageous from a handling and installation viewpoint. Since some degree of aerodymamic vibration of the towers is possible, it was considered imperative that all anchor bolts be pretensioned and that under sustained loading; elongation be held to a minimum. For each tower, 36 one-piece bolts, projecting 2 feet above the concrete, were equally spaced around the vessel perimeter. All bolts were threaded on each end and anchored mechanically at the bottom with
steel were considered
a 2/2-inch thick
rectangular plate held between two
torqued nuts.
Octogon Pedestol. Because vertical loads for both towers were considered equal they were centered symmetrically about the foundation centerline. The condition causing the greatest eccentric loading from vertical forces alone would result from either tower being water tested singularly. This eccentricity, found to be considerably less than that caused by maximum wind forces, was not critical. The combined wind overturning moment from both towers applied at the top of foundation was 27,750 foot kips. Any wind shielding effect by either tower was neglected and the overturning moment was assumed equal in all directions. Because of this, an octagonal outline for the foundation mat was more suit-
a square or rectangular shape. Using a maximum toe pressure of 4,000 psf, a 50-foot diameter octagon was found to satisfy all load combinations, with the number 2 load combination actually controlling the diameter. The weight of the operating liquids was relaively small when compared with the total able for limited soil pressure than was
mass and overturning moment. As a result, there was less than 10 percent difference in the toe pressure andf or
46
ELEVATION
FIGURE 3-Plan and elevation showing spaced
at 28 feet on
centers.
DESIGN CONDITIONS One tower is ll/z feet in diameter by 177 feet in height. The other is 10 feet in diameter and 203 feet high. Although the towers difiered considerably in size, there was less than 3 percent difierence in the calculated vertical loads for each. This was found to be true for both operating and empty conditions. For design purposes, vertical loads for each tower were considered equal. Empty tower weights included the vessels plus all accessories ready for operation, Op""ating weights consisted of empty tower weights plus operating liquid. Other data and conditions which governed foundation design are as follows:
Tower weight, empty, each 425 kips. Tower weight, operating each 500 kips.
Maximum velocity of wind, 125 mph. Maximum allowable soil pressure, 4,000 psf. Maximum settlement allowed, Minimum stability ratio,
/z nch.
1.5.
Concrete-3,0O0 psi in 28 days. Where applicable ACI Code (318-56) to govern design and detailing. f" and f" to be increased by one-third where stresses are due to combined wind and vertical load.
Maximum allowable anchor bolt Alloy steel, 40,000 psi. Carbon steel, 20,000 psi.
stress:
Unusual Foundation Design . . . eccentricity between operating and empty conditions. The stabiiity factor under load combination 1 was 2.4, and for load combination 2,2.6.
Step Seclion. After the mat had been sized it was determined, by trial and error calculations, that a stepped section through the center was desirable. This step was run continuously one foot thick, across the mat center and for convenience was made equal in width to the octagon side. At the edge of the step, the mat depth was set at three and a half feet which with sufficient bottom steel would approach a balanced design for the resisting moment from soil pressure. This depth was then continued to the outer edges of the mat in order to minimize the steel requirements and to maintain over-all stiffness. The pedestals were then made equal in width for symmetry and connected. A minimum allowable cover of 12 inches, outside of anchor bolts on the 11foot, 6-inch tower, determined the 14-foot, 6-inch width. Mqt Reinforcing. The mat reinforcing in the transverse direction or perpendicular to face of the step was determined by analyzing sections across the entire width of the foundation. Shear and moment at sections through the center, at the face of pedestal, at the face of step, and at points between the step and outside edge of mat were c;mputed. Load combination 2 caused a maximum moment at the face of the pedestal and step and at other points toward the edge of mat. Maximum moment in the same direction through the foundation centerline was caused by load combination 3. To satisfy this shear and moment, eighty-one ff 11 bars were spaced on six inch centers in the bottom of the mat at the face of the pedestal and step, forming a center strip 40-feet wide. Alternate bars of the above group, plus three shorter f 11 bars along each edge were extended through the center to the opposite side totaling +7 #Ll bars to resist moment through the cener. As the moment decreased toward the outer edge of mat, alternate bars were discontinued in two stages leaving f 11 bars on 2-foot centers at the extreme outer edges. In computing transverse reinforcing requirements for the top of mat, negative moment on the windward side caused maximum tension at the face of the step. Here, +l #tt bars were placed on one-foot centers with al-
Aboul ihe Author Edward V. French is a senior structural design engineer with Phillips Petroleum Company, Bartlesville. He directs the structual
and civil engineering design phase of assigned projects. Holder o{ a
B.S. degree in civil engineering from the University of Missouri. He has been with Phillips in the
Engineering Department since his graduation in 1952. Previous to this time he had two years' experience in general construction work.
in two stages both toward the outside and toward the center, leaving ff 11 bars at 4-foot to run continuously through the center. In computing this mornent, only the weight of the overburden directly above the mat, plus the concrete in the mat, was considered acting downwa,rd. ternate bars discontinued
The heaviest reinforcement in the opposite direction or parallel to the longitudinal axis of the pedestal was also required for load combination 2. Assuming the wind from a direction parallel to the longitudinal axis of the pedestal, tension from the wind moment on the leeward vessel combining with the effect of soil pressure produced maximum tension in the top of pedestal. This was near the inside face of the leeward tower. Assuming all the tension to be resisted by longitudinal steel alone, 22 f 11 longitudinal bars were placed for this purpose in the top of pedestal. For the same combination of forces, tension in the bottom of the mat near the inside face of the windward tower required. 47 ff10 longitudinal bars. These were placed in a 2O-foot wide strip through the center o{ foundation. Other reinforcement in the longitudinal direction was of a nominal nature and was placed in sufficient quantities to assure proper continuity.
It was first considered desirable to specify a continuous concrete pour between the mat and pedestals thus providing the best possible shear connection between the sections. Several factors making a continuous pour impractical were excessive dead loads on forms; Sheor Key.
inaccessibility; possible difficulty in positioning bolts; and unnecessary exposure of the excavation to weather. A large portion of the anchor bolts and pedestal reinforcing totaling some eleven tons would normally require support from pedestal form work and create a support problem. Concrete placement in the center portion of the mat would be difficult with all pedestal reinforcing and bolts in place. ft was felt that accuracy in positioning anchor bolts might be sacrificed if a continuous pour was made. Assuming a continuous pour, the excavation would be exposed to weather longer before pouring could begin, thereby subjecting the soil below the footing level to detrimental change in mois-
ture content. fn order to eliminate these disadvantages, a construction joint was designed between the step and pedestal that the mat and step could be poured first. A 14-foot, 6-inch wide by l-inch deep recess centered beneath each tower provided a four way shear key between step and pedestal. This recess also provided additional depth for maximum bolt anchorage. so
Vertical R.einforcing. Particular attention was given to the selection of adequate vertical reinforcing through the center of foundation, tying the mat and pedestal together, because of the unusually high vessel overturning moment. The two, fourteen and one-half foot octagons were first assumed to act as separate round stems and the connecting center section neglected. They were then analyzed as round sections acting in bending and direct stress, the critical section being taken at the construction joint. This analysis resulted in a total of 120 square inches of vertical bars required for each stem. IJnder this assumption, these stems could transfer all of the over-turning moment from the towers without influence
47
from the connecting center section. This connecting section became functional when the full depth of the foundation was considered a flexural member resisting a mo-
ment in the transverse and longitudinal directions. For vertical bars in each stem, 1'20 #9 bars were arranged into two rows, one row on either side of the anchoi bolt circle. It was felt that in placing these bars in two rows, stress from the anchor bolts would be transferred more evenly and that any tendency for the concrete to separate at the construction joint would be minimized. Additional ff9 bars were then spaced on l-foot centers along each side of the connecting center section to prevent separation at the joint when the entire foundation acted in flexure. Figure 2 shows the foundation after the first pour was completed and it also shows the vertical bars and the keyed construction joint used to tie the mat and Pedestal together.
Gin Pole Bases. The possibility of combining a
lt
r [,
con-
crete base, which would supPort and anchor the tower erection gin poles was considered. By providing such supports, considerable time and labor could be saved when setting the poles by eliminating the need for tying down the pole bases. The position of each tower prior to raising was planned with the tower lying at 45" to the main foundation axis. The pole bases at the closest possible position would straddle either tower on approximately 30-foot centers. Figure 1 shows the poles with the 10-foot by 203-foot tower free of the ground support. When the foundation was analyzed., applying concentrated vertical reactions from the poles spaced at 30-foot centers, it was found that tension across the top of the concrete might cause extensive cracking. This cracking, although probably not detrimental, was undesirable and to prevent it, additional heavy reinforcement would be required. The estirnated additional cost of materials to provide these integral foundations was estimated at $2,000. This was considered too costly for the advantages offered and the plan was abandoned.
;:K:i l|.::,.4
.:N
:M :,,# 1#/
#r
i, *
rl=
ffi el::ilNl
:;ag=
P#,,ffi
As an alternate method, the gin pole bases were set outward and placed on timber cribbing completely clear of the tower foundation with cables providing the necessary anchorage.
Leveling The Towers. The pedestals were poured to within 2 inches of the finished elevation. As the towers were erected, the base rings were set on a series of steel shims which had been previously leveled. Final leveling of the towers was then accomplished by adjusting shims and anchor bolt nuts. After all adjustments were completed, two inches of grout was placed across the top of pedestal and beneath the tower base rings. Each anchor bolt was torqued to an initial stress of 45,000 psi. No inconvenience was reported by the contractor because the anchor bolts projected two feet above the con-
crete. Neither was there any difficulty reported in regard to spacing anchor bolts to match the tower base
FIGURE
LTower
installation complete with insulation,
platform and piping.
two independent foundations. This was substantiated by further experience when a third tower of similar propor-
tions (t}/z feet x I77 feet) was designed and installed simultaneously, nearby. This tower was placed on the usual mat and pedestal octagonal foundation and required only 140 cubic yards of concrete. However, under the circumstances which established the design conditions, there were still advantages in the saving of space in conformance with the best piping arrangement and
in the possible vibration damping eflect gained.
rings.
Mqteriqls. A total of 383 cubic yards of concrete and 27 tons of reinforcing steel was placed in the foundation. As mentioned previously, no large savings, if any, in concrete materials were realized over those required for
48
Trend. There is a definite trend in the industry toward the use of taller fractionating vessels containing more trays. The experience acquired during the design and installation of these towers will be useful in determining rL JJthe feasibility and planning of future units. 1+1+
NOTES
49
FIGURE
FIGURE
1
FIGUR.E
2
3
Foundation sizing simplified Tables can be used to select foundations as easily as capacity tables are used to select pumps
/_t _7 v.__r_k, K s,= tt/64 (5-8k) I --6 ^ \lo --
**u')
I t L:r r8k-5) rr.rin(2ks2 ""' !k-*-r
1)
2(1-k)
E = D"X (2C'lC') '2r-',
Dovid H. Konnopell, Girdler Construction Corp., Louisvllle,
P
KY.
using JUST AS a designer can select a storage tank a structural designer can choose a forndaiion based on tables of capacities' Entering the tables with a given weight and moment (or eccentricity) you can q"iclly select the minimum size of foundation required. In addition, you can readily determine the dislribution and magnitude of soil pressures under the
capacity tables, so
The capacities of an octagon foundation subjected to loading ai sho*n in Figure 2 are determined from the following formulas'.
(1- m) E,a " D' B ' (1-l m) (t*m)l P:BE'zp X (1-*)' M: EP
foundation.
How lo Moke Copociiy Tobles. Two cases of foundation loading are considered in developing capacity tabies. The fiist case consists of a loading which produces uplift on part of the foundation. This case is shown i., Figrt"- 1. The second case covers bearing under the eniire foundation and is shown in Figure 2' For calculating the capacities of an octagon foundation subjected to soil pressures as shown in Figure 1, the following formulast are used.
c,:,r/B(t-2k) + (+-+- *+k, (2k-1)
r 50
4
)J k-k'+ arcsin
(2k-1)
: CrpD"' M: EP
NOTATION
P: M: E: : k: --
p
m
:
D. :
Concentric vertical load capacity, kips Overturning moment capacity, foot-kips Eccentricity of load to produce corresponding moment, feet. E,- M,/P'
Unit soil pressure, kips per sq. ft. Ratio of unloaded length of diameter to diameter of int".iU"a circle of octag6n. Used in Figure 1 loading only' Ratio of minimum unit soil bearing to maximum soil bearing. Used in Figure 2 loading only.
Diameter of a circle equivalent to the inscribed diameter, D, of an octagon, feet. D": 1.04D.
Cr and Cz are coelTicients rrsed to shorten algebriac
opera-
By decrementing k and incrementing m, capacities of a one-foot diameter octagon are developed based on a maximum unit soil pressure of one kip per square foot. The relations of the capacities and eccentricities of any other diameter octagon) D", to those for the one-foot [,,,_6,,
P1
0,,
Mr,-0,,
X D,'for
a given
Step l-Assume weight of foundation, pier and earth backfill as 50 kips.
X D*'for
f
:
SO.O
of
Tables 1 and 2 are illustrative of tables that may be used to estimate and design footings subjected to the forces described. The tables were developed on an elec-
tronic digital computer.
p: ''.A,djusted" "
1+7 100
:
190:o 2.O
1.47
-
ft.
5o.o kips
Step 3-Enter tables with "e" and "adjusted" P. Select ll'-0" octagon (Table 1) (table capacity P : 51.393 kips, "E" : 1.429 ft., k : 0.0) .
ILTUSTRATIVE PROBTEMS
Step 4-Check assumed foundation weight:
To demonstrate the use of the capacity tables, several illustrative problems are presented as follows:
Foundation: 11'-0" Oct. 1'-6"
Problem l. Size an octagonal foundation for the selfsupporting vertical vessel shown following data:
p:2000
: M: P"
lbs./sq.
ft.
in Figure 3, using
the
(2.0 kips,usq.ft. at 1'-0" below grade)
50 kips, weight of tower 147 ft.-kips, about base of tower.
Since the tables show capacity of foundations based
on a maximum soil pressure of 1 kip per square foot, it is necessary to first use an "adjusted" value for P and M to compensate for the larger soil bearing value. The following relationships are used: "Adjusted" P:
Allow. pressure, kips/sq.ft.
Pier: Backnr:
\23
1.15
l:dB
.00
/.05
I.ro
85.596 84.784 83.016 80.412 77.147 73.+93 69.818 66.1 43 62.469 58.794 55.119
l.+s
(
.so l.oo
l/.ig \:33
64.242 66.812 69.381 71.951
5t.445 47.770 44.095 40.421 36.746 33.071 29.397 25.722
74.527 77.O90 79.660 82.230 84.799 87.369 89.939 92.509 95.078 97.648 100.218
Capacitles based on
1000
.542 .476 .415 .JD / .303
Ahernqte Solution. Step l-Same as in original
.158 .1
2'-6'
15
.075 .036
lb/sq. ft. allowable soll bearing
:
100.2 sq.
ft.
Thickness Wetght(ktps)
l,-4"
i:AB
.00
i.O,5
/.ro
r5.0 22.5 30.1 .J/.O
=##
ili:
\.ls l.ts
1 .so
\.ss 1.60
l.os
t.Il \
:BB
solution.
2-Octogon Diometer :
l4.OO
M 176.466
59.949 64.854 69.679 74.387 78.933 83.249
)74.793 777.147
i65.778 159.048 151.514 143.938 136.362 128.786
87.4t2
97.574 95.737 99.899 104.062
721.21,1 r 13.635 106.059 98.484 90.908 83.332
\o8.221
.204
3.674
Area of base
1,-6, 2,-o,
\:33 {.15
22.O47
18.373 14.698 11.023 7.349
,
10.87 kips
22.50 kips
Re-enter Table 1 and check selection. Inspection indicates 1l'-0" octagon is satisfactory. Since k:0.0, the distribution of soil pressure is such that 100 percent of the foundation is under compression; minimum soil pressure is zero on the windward edge and 2,000 pounds per square foot on the leeward edge.
TABLE
2.177 1.929 1.751 1.583 1.429 L.293 1.169 1.056 .953 .857 .769 .688 .612
high : thick :
(100.2-20.7)x.i.1io
korm 37.009 40.037 43.016 45.923 48.729 51.393 53.963 56.533 59.102
6t.672
\r#;
5'-0" Oct. 3'-6"
P: 50.0 + 53.25 : 103.25 kips 117.0 _ r.42 E_ 103.25 Adjusted "2 P - fq=! 51.6 kips
fABIE l-Octogon Diometer: ll.OO
korm
value
P: E
a given ,,k,, or ,,m,' value
100 kips.
and "adjusted"
Step 2-Calculate eccentricity
"k" or "m" vaiue
soil
Pressure.
p,:50.0
D* for a given ,,k,, or ,,m,,value
X
Allow. pressure, kips/sq.ft.
The eccentricity, E, remains the same for anv
octagon are as follows: f,- : P, : M" :
M
M:
"Adjusted"
tions.
112.386 116.549 1,20.712
r21.874
75.756
12S.036
68.181 60.605 53.029 45.453 37.878 30.302
133.199 137.361 141.524 145.686 149.849 154.01 1
22.726
158.174
15.150
7.575
162.3116
Capacities Based on
1000
2.943 2.695 2.456 2.228 2.O74 1.819 1.646 1.489 1.345 7.273 1.091
.979 .876 .779 .690 .606 .528
.454 .386 .259
.202 .747 .095 _046
lb,/sq. ft. Allowable Soll Bearlng
162.3 sq. ft. Thickness Wetght(kips)
Area of Base
1,,-01 1',-6',
2'-0, 2'-6'
24.3 36.5 48.7 60.9
5l
Problem 3. This problem illustrates use of tables to determine soil loading under an existing foundation. For this problem, refer to Figure 3 and use the follow-
Foundation Sizing Simplified . . Step 2-Obtain "adjusted" values of P and M: "Adjusted"
": ff -
"Adjusted"
Step
ing data:
'B:60: 50.0 kips 117:0: 73.5 ft.-kips
P": 300 kips, weight of tower M - 373 ft.-kips D: 1l'-0", diameter of octagon
2.0
3-Enter Table 1 with
these adjusted values. Se-
lect 17'-U' octagon as before. Step 4-Same as
Problem
in original
2. This problem
solution.
illustrates the method for
obtaining sizes of foundations other than those given in the tables. For this problem refer to Figure 3 and use the following data:
p:5,000
lb.,/sq.
ft. (5.0 kips,zsq.ft.) at4'-0"
E 235 kips, weight of tower l![: 990 ft.-kips In the solution of this problem, the following relaP,
In the solution of this problem, the following relationship is used: p, _ proro,:j fo" t given "k" or "m" value' 1.0 P P1 When pr : actual maximum unit soil pressure, Pr: total vertical load, P : Table I octagon load capacity based on i.0 kips per square foot soil pressure. Step l-Calculate weight earth back-fill:
5'-0" octagon 3'-6" high Foundation: 11'-0" octagon 1'-6" thick Back-fill: ( 100.2 20.7) x 2.5 x 0.1
Pierr
tionship is used:
#:+ t'k"
for a given
e1!,:
{#r
-
"'
Pr : P:
Total load
'
load to be carried by octagon D1 table octagon load capacity.
:
Step 2-Calculate
of P:
"r. -
+
235.0
85.0
:
320.0
:
3?o:o "Adjusted" P" 5.0
3.09
-
p.
value
ft.
64.0 kips
3-Enter Table 2. Closest capacities are P : 59.949 kips, E :2.9+3 ft., k : 0.25 for l4-0" octagon. P : 68.819 kips, E : 3.153 ft., k : 0.25 for l5'-U' octagon.
Step
D:
E: Step
{
640/59.949
14.5/14
X 2.943:
,[-Check
3.06
M.47 ft., say 14.5 ft., and
ft.
5'-0" octagon 3'-6"
Foundation: 14'-6" octagon 1'-6"
e
-
.9,9,0'9
323.47
32J'=47 p "Adjusted" -
5.0
-
3.06
: : -:
10.87 kips 3g.20 kips 38.40 kips
B8.47kips
ft.
:64.69
kips
D: { 64.69/59s49 X 14.0,: t4.62rt. Since k : 0.25, it is immediately known that 25 per-
cent of the diameter of the octagon is unloaded and 75 percent is loaded; the unit soil pressure varies from 0 on the windward side to 5,000 lbs. per sq. ft. on the leeward side over the loaded length.
52
-
300.0
373'0 353.25
:
+
53.25
1.06
-
353.25 kips.
ft.
:
:
-^814 59.102
5.98 kips,/sq. ft. ( 5,980 lbs. per sq. ft. on leedard Ldge.'
),
LITERATURE CITED
Fork, Chas. A., "Graphical Methods Aid in Stack Foundation
sign" Petroleum Refiner 30, No. 3, p Bf (1951). 2 Fork, Chro. A., "Applying Graphical Methods Design" Petroleum Refiner 31, No. 11, p 145 (1952).
to Square
Abouf the Author David H. Kannapell is a senior
with Girdler Construction Corp., Louisville, Ky. where he performs civil and strucstructural engineer
tural design of gas processing
and
chemical plants. Holder of a B.S. degree in civil engineering from the University of Louisville (1936), Kannapell has had structural de-
iI
sign experience in many large i chemical plants, synthetic am- [,.llii monia, hydrogen production, gas purification urid .*bid" rnurr,rfi.- D. H' Kannapell turing plants. He is currently the president of the Louisville chapter of the Kentucky Society of Professional Engineers.
De-
Footing
assumed foundation weights:
high thick Backfill: (174.1 20.7) x 2.5 x 0.1 Total P 235.0 + 88.47 : 323.47 hips Pier:
E
:
Minimum soil pressure ffipr : 0.15 X 5980 : 897.0 lbs. per sq. ft. on windward edge. 100 percent of footing is under compression. The method outlined herein has been limited to octagon foundations for brevity. Using appropriate formulas2, the same method may be applied to square foundations with -an overturning moment about both the rectangular and diagonal axes. 1
X t4xf :
10.87 kips 22.50 kips 19.88 kips :53.25 kips
3-Enter Table 1 for 17'-U' octagon with known "E." Read P : 59.102 kips. m.: 0.15.
320.kips.
eccentricity and "adjusted"
990'0
: : :
Step
Step l-Assume weight of foundation, pier, and earth backfill as 85 kips. P
Total
and
Step 2-Calculate total load and eccentricity:
or "m" value.
D, : octagon diameter desired D : table octagon diameter
Where
of pier, foundation,
Dowel Sizing For Tower Foundations
Tower pedestal dowel bar reinforcing is usually oversized by u commonly used formula with a high safety factor. A more economical method is presented
Andrew A. Brown, Union Carbide Chemicals South Charleston, W. Va.
Co.,
Trre' srzrNo oF TrrE DowEL reinforcement is usually the last part of tower foundation design. All combinations of loads and moments are required in computing the base slab. The pedestal size is usually fixed by the base ring and anchor bolt spacing. So, with these data, the size and spacing of the bars can be assumed for analysis. Some designers use a minimum percentage of the area of concrete for the reinforcement similar to concrete column design practice. Exomple. As an example, the following design data of an existing column will be used:
M:
7,000,000 inch-pounds, the maximum moment
K
o<
t.00
t800
.95
154"
09'
.90
t43"
08'
.85
134'26'
.80
I
260 52'
.75
t200
.70
I
t3"35'
.65
I
07028'
.60
tot" 32'
.55
95"44'
50
900
.45
84" t5'
.40
780 28'
.35
72.32'
.30
6stzs'
.25
600
.20
53" o8'
.t5
45"34'
.t0
cM-.
.20 cv -
.30
30
.03
.04
2.0 3.0 4.0 60 8.0 t0.0 .60 .80 t.0 .06 .08.r0 .20 .30 .40 .60 80 r.0 FIGURE l-Coefficient curves used to find rrnit stress.
20.0 2.0
30.0 400 -cM
3.0
3605 2'
4.0 6.0 8.0 r0. -cv 53
It is apparent that the latter solution is not very economical, and contains a factor of safety out of propor-
DOWEL SIZING FOR TOWER FOUNDATIONS
P
:
80,000 pounds, the minimum load which includes the weight of the concrete pedestal 36 inches, the radius of the inscribed circle
r: R: 33 inches, the radius of the dowel bar circle Reinforcement: 20 number B bars, NAs : 20 (.79) : 15.8 square inches n:10
With this information 11r"-e it computed, k values are assumed and the various determinations made until the neutral axis is located. Then, the -r-o1 4r" internal stresses equals that of the external forces. As a convenience in recording the values, a table is constructed. This can be revised to suit the individual. The analysis follows:
M Pr / R \'z: 2) Strr-2-lpnr (;) r
e r
\
3pn7:
3)
7,000,000
80,000 (36)
'79r / .;tr-\*/
2+ (10) (20)
3 (10)
z (20)
(.7e)
7/ (JoJ.
33
\'
-t.*t
:.Jbt)
--# Now try k : .26, from Figure 1, CM : 5.1 and CV : .41 (see table below for complete investigation which shows ef r : 2.01 or too small.) The other k values are tried until the efr approaches 2.43 ("k's" of -25 and .245 brackets this e/r). CM+SM Nore: SV : 3z'pn (l-2k) and ; : 16i6y_ g1z)
.26 .24
5.10 3.90 4.30 4.77
M+
sv
SM
SM
2.45 2.15 2.45 2.15 2.15
7.55 6.35
.48 .51
7.22
.50
7.O2
.51
e/r
CV 3.76 1.79
.410 .310 .340 .380 .360
.52
2.40 3_15
2.01 <2.+3 3.55> 2.43 2.81> 2.43 2.29 <2.+3 2.54> 2.43
The unit stress in the concrete can now be computed for these two k's by using formulas concrete, and
fs:
nf"
[R]r(1-2k)l
t: 1"#ffi
2kr
for the reinforcement. (Equations 5 and f"
ro.
96 (.25) 7,000,000 7.22 (36)3 (4e7) 10[33 + 1B]
:
:
497psi
f"
:
2 (.24s) 36
thL structure proportionally as he would never be retained for
a repeat performance. To stimulate and provoke thinking toward -the d-evelopment of a more rational analysis for dowel bars, this method is submitted. It is not presented as the final answer but with the ho-pe ihat it will influence others to produce something better for use.
For this presentation a cylindrical pedestal, or that formed bv the inscribed circle of thi octagon br other regular polygon is'used. The working stress design method is emPloyed with.the attendant assumptio;s. A section that is plane before bending
remains plane after flexure is imposed. Stress and strain vary as a straighi line and directly as the distance from the neutral axis' The re-inforcement takes all tensile stress due to flexure.
In the development of the formulas the reinforcement is rewith an-area of Es,/Ec times that of the steel. In conitructinq the transforrned section, the holes in the concrete were
olaced
not
remloved
.511
f":
:14,600
psi
By comparison with the conventional method of 4MP ND-Nrweget the force imposed on the maximum 80,000 4 ( 7,000,000) :21,200 4,000 : stressed bar : -lO-G6t- 20
derivations considerably. The svmbols used are the same as those usually found in concrete'design manuals and text books employed for teaching
this subject.
Fisure 2 shows a tvpical foundation with the forces acting on ii and sives the location of the dowels. A section is taken through thE pedestal just above the foundation slab and the forces"acting bn this section are located in Figure 3' The equations representing the total forces and moments im-
posed on the corr.r6t. and reinforcement are now derived' Taking the summation of moments about axis Y-Y we have M;- M" : O or M : M" + M". By summation of the M
in the Z direction we get P-(V"*V":O forces
P:
V" * V". The total vertical force acting on the concrete is the sum of all the stress acting on the segment of the circle to the right of the neutral axis.
If f"' represents the intensity of stress on the elemental area dA., then dvc - f"'dA", dA" : !16* : 2rsind (rsinddC) f .' : (cosd (1
-
cosa)
-
cosa)
54
stress is
By substituting
these values
total force acting on the concrete, V"
a) sin2ddd,
obtain V"
#:21,700
psi.
f"r2
-kL3l-sin3
in the above,'the
: r.:*rr l!*"r -
a
-
stn a cos-
a _(1cos 2
I
rve
, Equation
1
Taking moments of the internal stress in the concrete about axis Y j Y, it follows that dM" : xdV". The moment of the force f"'on elemental area dA" about Y-Y becomes dM": fo' dA" rcos d as
f"'
x:
rcos
d. Substituting the values of dA"
as before, the total mo nent becomes
M-"
2f - r3
(1
17,200 pounds.
The unit
from the compression area. This should have very
little influence on the end iesults and is partially neutralized by th" u."u outside of the inscribed circle. It does simplify the
Integrating and substitutins a for / and 2k for (1-cosa)
96 (.245) 7,000,000_ 502 psi 7.02 (36)3
(502) 10 i33 + (36)
this purpose. As expressed rffi-*ror by many, it is safe. Actually it provides a factor of safety out of propoition to the other designed elements and is merely an exp'edi6nt. One would not darJ to oversize the other parts -of uneconomical formula
cos
14,100 psi
1B
f.:
Derivqlion of Equolions T'he subject of foundation design for tall stills and towers during the past year as evihas been accorded much thought denced bv numerous articles. Other equally important items srrch as anchor bolts and dowels have been of less concern. Most writers subscribed to the use of the approximate, inaccurate and
f"
7)
use
should be discontinued.
our
2.4.\
-
tion to the other elements of the foundation. Its
"
2f"
: 11
-
cos
a;
l!*'r
-g6s
rs L,^,,. (ti t"' "1 l- "'8
sin
a)
sin2
+c)
o cos o d o cososin-Cl"
-
3
-l'
and
f".'
-
l-a
KL
f
cos a sin
a-
2 coss a
Equation
al
cos a sin3
3-],
B
2.
The total force V" acting on the steel is found by converting _ tle dowel reinforcement into a,n annlrlar ring of equivalent area of concrete and of width t- The width i. .{"ui t,i tt p..a""i of n and total area of the dowels divided Uy) 7n. L.;" i"-a;;i to thq intensity of stress acting on an area dA" which irfo.iiJ y. a distance of R cos 6 from axis y Then dV" : f"dA" : f" t Rd- d By similar triangles
f" R cos+-r cos a t - l(1 - cos ")
,,,
By substitution dV" : ,:,,2f"t\ "
-
f" (R cosd-r
(R cosd
r(I-cosa)
cos a)
r(1-cosa) r
-
cos a) d p
,,""t* . ("t' ,*cosc-rcosa)dp r(1-.cosa) ,)o
then v*:
"
: r(1-cosa) ,,2f"** .[*.r, d- -'"""* re.or*l l_ J., 2f" tR _ /_ 7r cos a) since A" : pt T 12 :2'n Rt - r( 1 cos a) (' 2 Rt - pnrz and (1 - cos a) : 2 k then f"P"t'J J"Y, v" : Equation 3. "2k -
The moment of the forces acting on the dowel bars about axis Y Y can be obtained by gett"ing the su--ation of the mo_ - of the forces acting oir all-the ments small dA areas. dM": dV" (Rcos d)
PE DE STA L A
-Jo FOUNDATION FIGURE 2-Typical tower foundation showing dowel trons.
loca-
2f tR: (R cos 9i- r cos a) cos p d / r(1-cosa) -ly- :,2!tR1 ('t'",*cosd-rcosa)cosddd ' r(1-cosa) )o :
:
f^tR2 l- n , ' ' ,'--R r(l-cos")L: 2
Substituting the limits, Ra
M"
: p;t
-l18oo
sinpcosp- rcosasinO ""*""'-Jo I
, and 2k
- ETY,
:
(1
-
cos
dA5 =
1
P66
s.
a) we get
,6
N il
Equation 4.
N
dA.=
About the Author Andrew A, Brown is a structural engineer with the Union Carbide
Chemicals _ Co.,
South
2tzt,nz4r*
(2ydx)
r(cos{-cos*)
Charleston,
W. Va. His work at Carbide includes the preparation of structural designs, repons and analyses for all types of frames and ioundations both new and exisring. Mr. Brown,s
Mc+ Ms
q*v;=e {-rcos*
professional experienie includes that of a bridge consuhant with 12 years
active dury Brown
in the U.S. Navy Civil
Enginecring Corps as a publii works officer and 10 years in rhe Bridge
Dept., State Road Commission if West Virginia. He holds a B.S. de_
gree in civil engineerin.q from the Universitv of West _Virginia. He is a member of thc Sociery o['American N{ilitary Engineers and Tau Beta pi.
SECTION
X.X
AXIS
FIGURE S-Section A.A through Figure foundation slab.
I
pedestal just above
55
l-Cotculoled Yqtues of numericol coeftitienlr
TABLE
DOWEL SIZING FOR TOWER FOUNDATIONS
Now Equation I and 3 are added and multiplied by "."
36'25',
r(V"
f
V")
.11
a
f"r3
sinacos2a-acosa
-k\3/sin3 : f"t'
1z sin3 a
,
f"pnrS '2k)
zr cos
7a.
25', 840 16'.
a\
90" 95" 44',
*
3
sinacos2a-3dcos
-3
pnz-cosa)
13'
720"
35',
'18.85
22.03
25.r2 2A.O3
30.65 32.93 34.79 36.19 36.92 d/.Dt 37.70
5.r0
5.87 6.71 7.58 8.49 9.42
1260 521 134" 261
r43'08/
151'09/ 180'
This is the product of r and the total stresses in the concrete and reinforcement and equals the external load P x r. The total
7.O5
9.67 12.58 15.77
2.50 3.06 3.68 4.36
7010 32', 7070 2a' 1
2.St
.22 .38 .59 .85 1.18 1.56 2.00
600 660 25',
:
.57 1.51
.04
45" 341
53'08'
Gil
GV and
moment of internal stresses is M. * M": Equation 2 *4
--k :
f"r'
3
t
(a.{
f"f3 96k
*
[rr,"
M:
e_
Pr
I
cos a sin a
-
cos a sin a
_
2 coss a sirr a) 24
2 cos3 a sin
a)
-
-
B cos a sins a
32 cos
(v;+vJ.
(a
I
cos a sin d
l6(2
sina
af
sin3 a
I
f"t'
2 cos8 a sin a) 3 sinacos2
a-3
32
(2 sins a cosasins
acos
4k 2+pn
coss
*#["t"*cososinc-2
M"+M"
6k 12
c
I
f"pnrR3'lr
+
*
(+);l
a sin a)
3 sin acos2 a
The observation is made that for any value of k or a, CM and CV can be computed. Tabie t has been computed for the values of k of .10 through 1.0 and the respective angles are
-c
Then, the unit stress
(cM *
in the
nf. [R f ,( ]
2tt -
SM)13
reinforcement is found as f.
2k]],
Equation
p: n:
24
pn
(+)'"
d-3 a cos a)
6.
32 cos asin3 a,
pounds at the bot-
tom of pedestal (section A-A) ratio of area of steel to area of concrete ratio of modulus of elasticity of steel
to that of
the
concrete radius of concrete pedestal in inches radius of dowel bar circle in inches external moment at the section total vertical force in the concrete
r: R: M: V": V" : total vertical force in the reinforcement (dowels) M" : resisting moment of the concrete M" : resisting moment of the reinforcement f" : ma*imllm unit stress in concrete in pounds per square inch
7
: diameter of circular pedestal f" : maximum unit stress in the reinforcing d
NOMENCLATURE
N: number of dowel bars A" : 6a", of one bar in square inches D : diameter of dowel bar circle in inches P': minimum total of vertical loads in pounds at the juncture of pedestal and concrete slab (section A-A)
56
pnrrCos
f
M: maximum bending moment in inch
noted.
using these values, the curves on Figure 1 were constructed with k and a as ordinates and CM and CV as abscissas. By the use of Equation 6, the neutral axis can be located. This is done by assuming various values for k until one is obtained that approximates the I of the external forces. IJsing r the curves, this determination is rather easy to obtain. The unit stress in the concrete is found by Equation 5; 96Mk . _
32 cos a sina a
, Equation
cosa
CM+SM whereCM:72(a cosasina-2 cos3 a sin a):---------i-* 16(CV SV) /nY SM : 24pn ( : I r, CV: 2 sins a* 3sin acos2 a-3 acos4 \r./ and SV--3pna.cosd
-
-3
/R'' a{24r" (:)
a-3 pnrr
Equation 5.
steel
in pounds
per square inch
2kr: the distance to the neutral axis measured along a radius from the point of maximum stress in the concrete. (kd) 2
subtended by radii drawn from each end oI the chord which forms the neutral axis
a: the angle
e: eccentricity in inches of M/P
!!
2,000,000
r000
6
900
o r
.il8
Ei
a E
L
t0A
G
= oe8/ =
G
=
N
.9
r o
I
N
@
&
.= 800 o
o
o F o .=
700 600
=;
500
c
o
o
=
a
=
=o
!.
a
800
r
600
*j
500
-
.9
J
B
-
700
400
F 60 50
--.e
40 30
4
20
o J
I
= E
o J
=
o
r
F
E
.9
o
=
.=
F
=
o
o
E
&
= J J
-6'o
r00
90 80 70 60 50 40
,30
FIGURB
l-This
nomograph calculates the overturning moment and the unit soil loading in tower foundation design.
Short Cuts fo Tower Foundation Design Graphic solutions to unit soil Ioading and Ioading caused by the overturningmomentwill speed up your foundation calculation time J. F. Kuong Atlas Powder Company Wilmington, Del.
"FOUNDATION DESIGN For Stacks and Towers,, by V. O. Marshall was the subject of a special supplement published in the May 1958 issue of parnoijurvr RorrNEn. This article, in turn, supplements Marshall's article in that it presents two nombgraphs for short-cut m.ethods- to the analytical formula tichniques requiring trial-and-error calculations.
.Irr designing foundations for self-supporting towers, with respect to the supporting soil, two-main considera_ tions are taken into account I a) the unit soil loading, and b) the tower stability. These two factors must be studied. The first so that the maximum load the soil supports will not be exceeded and the second to prevent overturning of the tower by external forces, such as those caused by the wind pressure acting on the tower. Calculations for this type of foundation requires a trial-and-error procedure. The size of the foundation is assumed. Then, the soil loading and stability are
57
Eq
50 40
i
E
r00000 90000 80000 70000
Where.
60000
S , Con Be Either
tp00p00
50000
S, 0r S,* And W
e0q000
ls
800,000
W
K,ls
700,000 Q)
0r Wr,
. 40000
Respectively.
The Areo Shope
Foclor. d, ls The Short Oiomeler 0f The Bose
600,000 A 500,000 (l)s@
4OO,OO0
uotion
E
il A -.\ -ci J
.:
O o E
= o
oC
.:F
3OO,OOO
E
.9
o 200,000 t o E
-^ -o
=
J
9
I
u')
c)
€)
t00,000
o a
e0p00
o)
C
80,000
€(,
70p00
O
o J
EN '6
= =
$
r0000 9000
- --
8000
.1000 Squore
3ornnon x o ---/--f-.azs
o .= J o C o o
=
E
o
go LI
6
o
3 J
o
7
5
20000
a
t0 "E o)
30000
o
Ngo!-uelsson ;= .7854 Circle : E
-E \<
G.
=
5000
=
4000
E' E' o J
60,000
6000
C
rrl
t
7000
50,000
:o
40,000
+ C
3000
cn
=
2000
30,000
r000
20,000 Nomogroph No. 2
900 800 700
Key:W+d+R-K.-S
500
10,000
FIGURE 2-Use this nomograph to find the minimum and dead soil loading for tower foundation
checked and used as
a criteria to determine the
suit-
abiiity of the foundation size originally assumed. In estimating the maximum soil loading, two kinds of loading must be considered, namely: a) the unit soil loading due to the dead load (which includes the weight
58
600
design.
of the empty tower, appurtenances and foundatioli, as well as the earth fi1l on top of the foundation base) and, b) the unit soil loading caused by the overturning moment produced by the wind or any other lateral forces acting on the tower.
it is erected and empty and does not include auxiliaries. As explained in more detail by Marshall,l in calculating the stability of the tower, the maximum soil Ioading must be used in equation (7) as defined in this paragraph. Calling the minimum soil loading S'- and Wr the minimum deadload, we have: when
The total soil loading is, therefore, S:Sr*So where: S
:
Total soil loading,
psf.
: Unit soil loading, dead load, psf. So: IJnit soil loading, moment, psf.
Sr
S, and
So
s.*:#ft-
are calculated as follows:
(2) and Sr: Sr:-i-wM" t
(3)
where:
a. is the area of thc base of the foundation, sq. ft. Orertrrrning moment about the base of the foundation, Mr, - foot-nounils. W.- is th'e rvcight of the empty tower plus the weight of the foundalion itself, including the earth fill on top of the basc (minimum dead load), plus the weight of auxiliaries to include the weight of the tower contents and appurtenances, in pounds. Z.' is thc'icction nrodulus of the basc of the foundation which varics with the geometric shape, cu. ft. Now, since
a:K(d)2
(4)
Z:F(d)3
(5)
and,
where d is the short diameter of the foundation base and K and F are proportionality constants for a given geometrical shape of the foundation base (octagonal, round, etc.), and since for cylindrical towers the overturning moment is given by (6) Mr : 0.0025(v)" Do'H'L equations
(2) and (3) can be written
as follows,
^w u1K.d2 so:
(7)
and the condition of a perfectly balanced system, as explained in mcre detail in the reference article, is
Sr-:
(
10)
and for an actual system Sr- should not be less than
So.
(B)
So
Nomogrophs. Based on equations (7), (8) and (9), two nomcgraphs have been prepared which reduce the time required in repeated trial-and-error calculations. The first, bascd on equation (B), gives directly the valtie of the unit soil loading due to the overturning moment, M1, when V, Do, H and h1 are known. The second nomograph solves both equations (7) and (9) when W, Wr and d are known. The nomograph is the same since equations (7) and (9) differ only in the value of W which is required to calculate Sr or Srm. Exomple. Consider the same example given in the Marshall article (to which reference is madc for detailed calculations) and compare the solutions obtained for 51, So and S,,, using the nomograPhs presented here, with those given in the original reference. The following data are given: Tower diametcr inc. insulation, Do : 4.5 ft. 30,000 lbs. Weight of empty tower. . Weight of assunrcd concrete foundation volume based on octagon-shaped
0.0025(V)" Do'H'L F.de
(e)
Weightof earthnll....
base.
Minimum dead toad, Wt, (30,000 + 63,000 +
63,0001bs. lbs.
..32,700
......125,7001bs.
3'.2,7o0) cti.,plusliquid.
Weight of auxiliaries, insulation, platforms, piping,
Here,
V, wind velocity,
mph.
Total height, FI, 54 ft. Height of foundation, hr, 6 ft.
Do, diamctcr of tower including insulation, ft.
H, height of tower, ft. L, lever arm of wind load, in feet, calculated
L:h.+Ll12.
as follows:
h1, height of foundation, ft.
when the tower is installed by itself.
In
otl-rer words,
Aboul the Author J. F. Kuong is a process engineer for Atlas Powder Co., Wilmington, Del., where he works in process improvement, trouble shooting and cost reduction. He is currently in charge of a technical section doing technical-economic studies, process improvement work
and technical support for line supervision. Holder of a B.S. degree
in
chemical engineering from the
in
of
San Marcos, Peru,
chemical engineering from the University of Pennsylvania, he has been with Atlas since 1954. Kuong worked in the Technical Department, Atlas Point Plant until 1956, when he became technical assistant to the production superintendent. Kuong has been a process engineer since 1957.
and M.S. degree
lbs.
Assumed short dianreter of octaglon-shaped based, d, 13.5 ft. K, arca proportional;ty constant for octagon base is 0.828. F, section modulus proportionality constant for octagon-shaped base is 0.1016.
Furthermore, the condition of poorest stability occurs
University
......48,500
Do'H : +.5'54 : 243.8. 12: 33. Enter 243.8 on D"H scale (Figure 1) and align with L : 33 on L scale to To calculate S", multiply
Calculate
L: 6+
54
intersection with first reference line. With reference line as a pivot, align pivot point with V : 100 to obtain Mt : 200,000. Connect 200,000 on Mt scale
with reference point for octagon section modulus factor and second reference line. Finally, align pivot point on second line with d : 13.5 and read So : 800 lbs./sq. ft. on extreme left scale. The calculated value given in the reference is 803. To calculate 51 (use Figure 2). Align W : 174,200 on left scale on nomograph No. 2 with d : 13.5 and reference line. Connect pivot point on reference line with octagon area lactor and read Sr : 1150 lbs./sq. ft. The reference article gives 51 : 1155. To calculate Sr- (use Figure 2). Repeat procedure
outlined just above except use W1 : 125,700 instead of W : 174,200, and read S1m : 830. This value compares l
with 830 as given in the reference.
Marshall,
V. O.,
Prraomuu Rrrrrrn,
LITERATURE CITED Foundation Design Handbook for Stocls 37, No. 5, Supplement (1958).
md Twes,
59
SSEL FOUNDATIONS
*sw
i-'s
Foundafion Design for 8-Legged Vessels
Using one general equation for bending moment, the reinforcement bars for the entire foundation can be calculated A_ndrew A. Brown, Olefins Division, Union Carbide Corp., South Charleston, W. Va.
TrrB rouxoerroN for the SJegged cylindrical vessel in Figure 1 can be designed with one general
shown
equation. Because of the relatively low height of these vessels, the stresses at full load are usually not influenced by
unit
wind or seismic forces. That is, when the allowable unii stresses are increased by one-third for loading consisting of combined maximum vertical and horizontal forces, the elements of the foundation are not usually overstressed by such loading.
CONCRETE
GRADE\
\
CONCRETE
SLAB
Bose Slob. The base slab for the foundation is octagonal. Formwork for this shape is less costly than for a circular shape, and the distribution of stresses is more uniform than for a square. The base slab is assumed to be divided into four equal bands as shown in Figure 2. This is a view looking up from underneath the footing. The outlines of the overlapping bands form soil pressure prisms. One of them is included in all four bands, six -are in three, four in twq and two are in one band only. Since all four bands are identically loaded, one will be
removed and treated as an independent simple beam span. The reactions are the pier reactions. Section A-A (Figure 2) is formed by a plane passing through the center of the piers. The magnitude of the loads or soil pressures have been drawn to a vertical scale to show the fraction of the uniform load that is supported on the span. If the load prism is in all four bands, the load intensity is one-fourth of. u. If. the load prism is in three bands, the load intensity is one-third and so on. - If we let D equal the short diameter of the octagon, in feet, and P the total load or soil pressure on it (excluding the weight of the top fill and the concrete slab) the uniform load becomes P/A: P/O.BZBD, : w in pounds per square foot when P is in pounds. The reaction for the beam is P/8. With these loads, the table of areas and moments is
PiA
=;T-
EAR HEN
f,__
FILL D
Fig. l-Elevation of typical
vessel and foundation.
computed (Table l). The last column gives the moment of_ the respective load prisms about the center of the span. The values in the other columns are labeled and are selfexplanatory. The total moment is 0.03282oD2 for one band. The bending moment on a width of beam of one foot is
*
_ *, _ O.\Z2BwDa 0.414D 0.414D : 1.25(D /2 - x) - O.o795DluD
_0.1035uD2 (D/Z
where " x" is the distance from the outer edge of the octagon to the center of the pier or reaction.
- Since two-way reinforcement is to be employed, the influence of the two bands which cross this one at an angle of 45 degrees will have to be taken into account. A section of unity width is removed from the center of the span and the value of moments imposed at the center by these bands are shown in Figure 3. Let m: moment on the main band acting on unit 6t
FOUNDATION DES]GN FOR S.LEGGED VESSELS I I I
i i
I
t
I
,
60
I
.-'L
SPAN
t I
:
j g
l
{
I t
L**
-*---*"c
$
Fig. 3-Moments about section of unity width.
IOP OF
width. Then the width at a 45o angle is 0.707 and the moment is O.?07m. The total moment for the two bands is 2 (O.7O7m) : 1.41m, but these ate at +5 degrees to the main band. Accordingly, the component to be added is m(1.414 X 0.707) : ?n and the total moment is 2m' For one foot width of beam, the bending moment at the x)- O.0795DlwD center becomes M : 21.25(D /2
FOOTING
: lI/2(D/2- x)-0.159D)wD
_ P "-A - D2x.828
..._ P
to exceed 350 psi.
The pier reinforcement is computed by the usual formula uied for column design. The minimum steel requirement would normally govern. The maximum tensile stress in the pier rebars is obtained when the vessel is empty and maximum wind or seismic forces are imposed thereon. The concrete is under maximum stress when the vessel is full and all other loads are applied. If the octagon is much larger than the outside to outside distance of the piers, a section should be investigated at the plane of the outside pier edge. Reinforcement would then be required in the bottom of the slab. This projection beyond the pier reduces the bending moment at the -L++ 1+ 1+ center of the span.
SHORT DIAMETER OF OCTAGON, D , FT. DISTANCES TO C.G.
43096 Dr 3r 904 D lI 2357D mo r6667 Dlll b :09048
D
E
w/4
Q sparu S
ECTION A-
(General Equation)
With this moment, the required reinforcement can be computed. In sizing the bars one should keep in mind that the steel is located near the top of the slab and the permissible bond stress is 3.+ Vf ilBar Diameter but not
A
SHOWS LOAD DISTRIBUTION ON THE BANDS
About the quthor
Fig. 2-Base slab plan looking up underneath the footing'
IABLE
Sec.
l-Toble of Areos
ond l[omenls
partment, State Road Commiss'i,on of West Virginia, and he hcts seraed as a,
Area ol Prlsm Base
I-
-042893D2
II.,.
.077067D2
II-a
.oo7359D2
III-b IV
.074719D2
Totals
-207704D2
I
ANnnow A. Bnowx, Captain, Ciui.l Eng'ineer Corps, [].5, Naaal Resemte, is a Senior Engineer, Olefins Di,ttisi,on, Uni,on Carbide Co,r'p., So. Charleston, W. VaMr. Brown's prof essi,onal enper'ience i,ncludes seaeq'al gears in the Bridge De'
.o71067 D2
Use
R:
.1035D22
Use .0323D2p
Note: The values in this table were extended further than the strength of the materials of construction and soil bearing determinations warrant. This was done
to check the work. For instatrce, the total area should eqval .4742D X .5D' and the total forces or reaction should equal one-eighth area of an octagon,
62
bri.dge consultant f or seaeral ciiies. Duri.ng his 12 years of actiue duty i,n the U.S. Nao'!/ some of his bi,llets useT e: Publi,c Works Officel', Naoal Ai,t" Station, Hampton Roails, Va., Naaal Aitr Stati,on, Kaneohe Bag, Hatuai,i, and Naaal Station, San'Juan, Puerto Rico; Design and, Constl'uctinn Officer, Fifth Naual Di,strict, Ma'intenance and Operations Oafficer, Eleaenth Naoal Distri,ct, and Assistant Publi,c Works Officer and Muintenance Superi,ntendent, Nat;al Air Trai,n' ing Bases, Pensacola, Fla. He is a, member of Intemtational Aisociation for Brid'ge and Structural Engineers, SAME, ASCE, and, has BSCE and CE degrees from West Virgi.nit Uni,t-tersity,
Pressure Vessel Foundation Design For vertical pressure vessels, the old middle third rule requires a safety factor of 3. These data show that a factor of 1.5 is quite in order J. A. A. Cummins Hudson Engineering Corp. Houston
IF THE FULLEST economy is to be realized in the design of a foundation, a complete understanding of its action under various loading is required. This is particularly true when the wind loading resultant falls outside the middle third of the foundation cross section and uplift occurs. In this case, the soil pressure, or pressure on the soil, varies
in a
Cose
different manner than when the resultant falis inside the middle third. It is emphasized that a factor of safety against overturning of 1.5 is quite in order, whereas the middle third rule gave a lactor oI safety ,_ of 3. The use of this data will result o-loin a precise, and hence more economical design and will be consistent with the safety factors derived from the American Standard Buildirg Code Requirements A58.11955, and the ACI Building Code. Advontoges of Squore Bose. For vertical pressure vessel foundations, a square base is preferable to an octagonal or round base mainly because of the complications involved in laying out the steel. An octagonal base requires at least three layers of steel, one on top of the other, and consequently a greater depth of concrete.
l-*f*
i
Cose
l-f_Cose
rrr
I
r_?..,
P,'co.a
Cose
f
II P2
Cose
III
r
rll,.
(
2.5
FIGURE
l-In
3.O
3.5
4.O
4.5
/YY - z"- Tu
Case
I tle
resultant is outside the middle third.
63
The base area is practically the same in both cases, but the octagonal base uses considerably more steel. Furthennore, the form work for an octagonal base costs more than a square one of the same area. A square base also permits iloser spacing of columns than the octagonal base.
The base must be set below ground, the bottom being below frost line and on undisturbed soil of known character. A pedestal is required to convey the load of the vessel to the footing. Ideally it should be circular, but an octagonal pedestal is cheaper to construct. The mini-
c'+ 2d
mum depth below grade of the top of the footing is often governed by the depth required to accommodate the vaiious pipes which are necessary wherever this type of vessel occurs.
Overturning And Soil Pressure. In any design, the factor of safety against overturning and the maximum
soil pressure must be computed first' The size of the base is then determined by trial and error. Figure 1 shows the relationship between the soil pressure p at the side of the footinf caused by wind normal to axis YY, and the soil pressuie p' at the corner caused by the wind nor-
mal
to
axis
and hence
with the eccentricity ZZ. The ratio $,raries p
with the factor of safety
about both axes; Case B A, 6
fl
r7.
The curve
r,r'hen the resultant is and uplift occurs for bending
shows three conditions: Case
outside the middle third
I,
:
II, when e", is less than
brrt there is still upli{t at the corner when bending
abottZZ; and
Case
III,
when bending about both axes
produces no uplift.
It can be seen that about T:2.8,p: p'. Below this value, p' is always less than p so that it is only necessary to investigate p. When 7 is greater than 2.8, the pressure on the corner is greater than that at the side. The curve is useful here to determine p' from the easily calculated value of p. It may be observed that the maximum ratio of
{will b"-l , : 0.84 p
1.19. For cases where 4 is greater
W
MX6V2
than 3.6, it is easier to comPute P' : * Ptus u:in the usual manner. It is usual when computing soil pressure to test for dead load plus test load with water and no wind, or dead load plus operating load plus wind load. Some designers still use dead load plus test load plus wind load together, but this is not necessary since the vessel is usually fi.lled with water only once or twice in its life for testing purposes. However, some companies advocate the
filling of
vessels with water when hurricane warnings are received, in which case the latter condition should be used for design, but this practice is not very common. For computing the minimum factor of safety against overturning, the empty weight of the vessel should be
used together with the weight
64
of the soil and footing.
z
Y
MN
FIGURE 2-The octagonal pedestal is reduced to a square of
equivalent area.
Allowclble Soi! Pressure. Allowable soil pressures quoted by most soil engineers are usually given as _a figure in pounds per square foot at a certain depth. This is the allowable pressure in addition to the weight of soil already there. Thus, allowance can be made for the weight of original soil above foundation depth when computing maximum soil pressure. This has a net effect of increasing the given allowable soil pressure by the weight of soil above the given depth' If this figure is used, allowance must be made for all backfill which may lie above the foundation as well as the weight of concrete and maximum weight of the vessel.
Wind Pressure. The American Standard Building
Code Requirements for Minimum Design Loads in Buildings and Other Structures, A58.1-1955, is the culmination of much study of wind pressures in the United States and contains recommendations for design pressures for different areas. It shows how wrong it is to take an arbitrary wind pressure and apply it to all Iocalities and to any height above ground as it frequently done today. Specification writers often specify a figure of pounds per square foot on a projected area which in many cases is too high for the majority of vessels, but which for very tall vessels, is actually too low. Also, they will sometimes specify a wind velocity and omit stating the height at which the velocity is to be taken. The Code gives design wind pressures recommended for any location in the U.S. and for any height above ground level. In addition, minimum recommended seismic factors are given and a schedule for recorded earthquakes in the U.S. with a rnap showing the locations of their epicenters.
I
Concrele Design. The concrete base should be designed according to the latest ACI Building Code (ACI 318-56). In Figure 2, the bending moment on the footing is calculated on Section MM and the diagonal tension on NN with bending about YY axis. The octagonal pedestal is reduced to a square of equivalent area and the length of the side C/ : 0.91 X dia. of an octagon. It can be shown that by turning the equivalent sqrlare
gg'-6"
J
through 45 degrees and investigating bending about the ZZ axts, the moment is never more than V2 X M**. Since the two-way footing will be designed for bending about YY, the components of the steel in the diagonal direction will be 2
4'-
L=5'-o"
\/2
The ACI Code at present makes no difference for a square and rectangular footing in the location of the critical section for bending. It is suggested that a more correct estimate of moment for square footings would be obtained by taking moments of the loaded area bcef about Section MM and distributing the steel computed over (ef plus 2d) where d is the effective depth of the base. This is particularly true with a small pedestal on a large base. The steel outside this critical section could then be placed at wider spacings. At present, the ACI Code specifies the moment to be taken and the steel to be distributed evenly over the full breadth of the base on Section MM. The diagonal tensile stress is to be computed from area bcgh on Section NN of width gh. The computation of bending moment and diagonal tension about these sections is best handled by use of formulae. These must vary according to whether the conditions are Case I or Case II (see Figure 1). Case III wiil be similar to Case II. The method of design is probably best illustrated by the following examples:
I T C'
FIGURE 3-Example Figure.
M": C': Sg: So: St: S.: D: d: B: {mto: 2or: en: Vo : V-: M-: S*: Mo:
70.84
X
4.25
Operating design load for
X 150 mat
: We : W* : W* : Wr : Wo : \{o"
is
studied
Earthquake moment bottom of footing, Ib. ft. Side of square of equivalent area to octagon, ft. Soil pressure for mat design, PSF Soil pressure due to total operating load, PSF Soil pressure due to total test load, PSF Soil pressure due to wind or earthquake, PSF
Depth of concrete, ft. Depth of steel in concrete, inches Side of square base, ft. Minimum FOS against overturning Operating FOS against overturning Operating eccentricity, ft. Shear at Section N, lbs. Shear at Section M, lb,/ft
Moment in footing on Section M, lb ft/ft Allowable stresses in accordance with ACI Code, psi Moment in footing due to uplift at Section M, lb inches
a09,620 lbs.
I Za,SOO
fts.
W" plus W,
:
438,420
I
45,100 lbs.
48t52olb'-
The following abbreviations have been used: M*r : Wind moment about bottom of footing at ]b. ft.
absorber
ft/It !o: Sum of bar perimeters in one foot width, S": Shear stress, psi f: Depth of soil above base, ft.
feet OD. 409,620 lbs.
typical
B
Texas.
EXAMPLES
empty
A
f I
with foundations located in Nevada, Oklahoma a nd Central
A typical absorber is taken, for example, with foundations to be located in (1) Nevada, (2) Oklahoma, (3) Central Texas (Figure 3). The vessel is 84 inches ID X 99 feet-6 inches S-S. The allowable soil pressure is 4000 PSI at 5 feet below grade. The shell thickness is 3.5 inches. Using 4 inch insulation gives 99 inches or 8.25
Weight of water to fi1l 244,500lbs. Maximum vessel weight 654,120lbs. Operating iiquid load Pedestal 9 feet-3 inches across flats;
I
t-
direction. Consequently, bending about ZZ does not enter into the concrete design.
weight:
6"
tX--:: : \/ 2 X the area of the steel
in the YY
Erected weight
H=l06'-o"
center,
Notes. (1) Unit weight of soil in calculations has been
taken at 90 lb/cu ft, being the probable minimum weight if unconsolidated. (2) For wind moment calculation allow 2 feet width for ladders, pipe, platform, etc., i.e.
65
totd OD vessel : 8.25 plus 2 : 10.25 feet. This 2 feet is conservative and may be reduced by an analysis of all the extraneous
,r"
-ffil*
r.,plusd,/6)(2x-y)
(Bx-r)]
y/B
plus
_2X483,520X4.13 _,e_jxTczr,
projections.
(l)-Nevada.
Example
Pressure zone. Wind pressure 0 on vert. proj.
-
30
-
50
30 50 100
100 feet
feet:
A58-1-1955 gives 15
shape factor
X
0.6:
feet: 20 ! shape factor 0.6: feet: 25 X shape factor 0'6':
up :
30
X
strape
factor 0.6
:
B: 19 feet-6 inches and D- 1.75 feet
Weight of base 19.52
X
1.75
X
Weight of soil above base 7t) 3.25 90 x (280 Operating load on soil Operating liquid load -deduct Minimum direct soil load Water to fill Max. test load on soil Soil pressure under test load
w"
483,520 99,820
5.54\2 483.520 M "-m : _)t-i_!----t_ttt| 58.5 X (16.2)'
M" :
x
lbs. lbs.
90,380 lbs.
wo: 673,720 lbs' : lbs' 6a45ZdiE; W-to: -28,800 W.: 244,500 lbs. lbs. w, : s89,42-o Wr
2,341 PSF
As Nevada is a region of recorded earthquakes, A58.1-1955, a seismic factor of 0.1 is taken'
see
58.5 plus_144,920 X 2.5) ^_^ 2,56+,750 plus 362,300 2,927,0501b. ftlft. This moment must be used since it is greater than M*1.
0.1 (+38,420
_ w.i,iX, --nfr,-_ _ woXB
?lmin
:
2y2,927,050
)
r.5
''n'- zM. _673,720y.19.5 -iX2,gfrrrc --2.24 _ 2,927,05O :
M.
e:"
673,720
wo From curve,
Io 12.8
and o4
)
1 i.e.
1V-_
r0.B
:
f":
4,000
:
36,700 lb.
ft,At
:45,100
lbs. 99,820 lb's.
12
M-
:
bottom: s* xf :1.12 in.z
steel in
12
X
36.700
z6r6ixo.B67x-r,
Use No. 7 @ 6 inches on centers
KF-
X
260
12
X
:75,300 > M
172
t2 check bond:
*'+-:
*
v* =ln_ C,)
v>.
The formula
B' g
at
Steel in
0.867
L39doo l l.0g
-
13,250
lbs./ft
x
17
X
5.5
"otil,rtjtnfffie and applies only when
(90 f plus 150 D)
toP
no compression steel
-
13,2s0
jd
4-135
zosoz
x
o
1 3 -1 \' \?-,, / (90
.s67
x
X
g.25 plus 150
(':'-tl;r
X f .i5) tb.tt/tt
X 12
bond
(90 f plus 150 D) (B
-
r7
x)
:
stress:ns6##X l^6 :
Wind pressure 0
30 feet 30
30- 50 100 50 100 up
on vert. proJ.
For?o
66
z
:0.13
in2
X
3.3
555
:
1,850 lbs.,/ft.
79 psi ok
PSF pressure zone.
f":
8.+2
B-C'
Example (2)--Oklahoma. A58-1-1955 gives a 40
20,000
( 3, proceed as follows: X:1.5 (B-2") :1.5 X 10.8:16.2 Y,: Yz (B - C' - d/6) - y2 (19.5 -
100
Use No. 4 @ 12 inches on centers KF > M le no compression steel
adding 33ys percent oversuess for ACI 603 (c) 26,667 1,500
f":
(
+ we i4402olb"-
V, :
<
85 PSI
(32.40 plus 5.54)
"
4.265 PSF
PSF
I
d) jd
:
t7
:-9:q
X 5: 450 Max. soil Pressure 3,815 Concrete design for 2r5(X) psi at 28 days 1,125
X
4.35 feet
Subtract wt. of original soil at given dePth 90
fo:
wb
M-.-u:
max soil pressure at side
673,720
_+X soilpressure-3s (B-2.) 58.5 x 4
Wp
: :
u=*l,,r=r?I::^. :2.15 -'-
2
(
,
:
plus 34) 0.867
u2W-"
:
4.13)l
168,480 lbs.
M-: - gn#-(2X plus u) where u : tt : /z (19.5- 8.42) - $.$4
B0 plus
Wo: Wr :
150
:
-
vo -n
.
(I2C'plus
168,480 ( 101
lb,4t
Let
4.13) plus l.2B (48.60
-
780 (159.08 plus 56.92)
S-:ShearStress: -s
18 PSF
10.25 (11,664 plus 60,000 plus 10,800 plus 5,400)
900,606
:
15 PSF
:
X
(8.42 plus 2.83) (32.40
12 PSF
10.25 (lB
50
l"
9 PSF
:
X 6 X 108 plus 15 X t2)(20X45plus9X30X20) M*r
a 20 PSF
-
2.83)
:4.13
X
shape factor 0'6
:
lB PSF
X shaPefactor0.6:24 PSF 50 X shape factor 0.6: 30 PSF 60 X shaPe factor 0.6 : 36 PSF
40
M*r:
10.25 (36
24)r.20 X 45plus
:
X
18
6
X
X
X
l0B plus 30 X 20)
30
X
50
80 plus
ShearstressS":# _
10.25 (23,328 plus 120,000 plus 21,600 plus 10,800)
:1,801,200 lb. ft.
ForB: 18 feet;D : 1.5;f - 3.5Weplus Wr : Wn: 70.84 X 4.5 X 150 - 47,820 lbs. *82: Se: 1,501 PSF 'Wo" Wu: Weight of base 182 X 1.5 Xl50 Weisht of soil above 90
x
3.5 (32+
load
deduct
Water to fill Max. test load on soil
wt/82
Soil pressure under test load Seismic factor 0.025 gives
486,240 lbs'
72'900 lbs'
-
79,700 lbs'
-28,800
638.840
X
18
2xl,Bot,2oo
:'t-19
610'040 lbs'
"-
for tt.:3.19 .r*. for4 p'"p'
2,637 PSF
652,000
lblft ( M*t
:
638,84t)
2,82 feet
{ :
0.914
\/ /\
I
:4.189
0.914
Subtract weight of original soil at given depth 90
N
pressure
(at corner) Max. soil
>
on-
:
''10' o
:
:
1,125 f"
-
5
:
PSF
450
-
Steel in bottom
:
1,801,200
Max. shear Vo
B2
i0
if S. < /s
Note: stress
:
56:
-
K)
S", then let
:
1.5
1,853 PSF
t3 s" (K plus 1) plus 2 s* plus K plus 1)
S*:
>
% Sr, compute Vo
(K'
l
0 and allowable shear
as above
and
S":
100 PSI
stress:
vos67d4:
Note:
il S* had
been
x
r'5ol
I'/s
11,664 8,175 plus 595)
-
34,77+
X
t2
:
14
ir2/It.
1.29
14,490
osoT
Se,
x t4X 5'+ - 221 PSI < 267
allowable bond stress
:
200 PSI
Example (3)-Central Texas. A58-1-1955 gives a
- 30 30- 50 50
X 25 X 20
:
factor 0.6
0.6: 30 X shape factor 0.6: 40 X shape factor 0.6:
100
-
shape shape
factor
12 PSF 15 PSF 18 PSF
24 PSF
M*r: 10.25 (2+X 6 X 108+ 18 X 50 X B0+ 15X20X45+12X30X20) : 10.25 (15,552 + 72,000 + 13,500 + 7,200) For
:
1,109,580 lb.
B:
16 feet-6
16.5,
X
1.5
ft.
inch D: 1.5 f:3.5 'Wo" : 486,240 - 82: S":
X 150 Wr:
1,786 PSF
61,260 lbs.
Weight of soil above
(272-71) W": Operating load on soil Wo : Operating liquid load90
X
3.5
deduct
Water to
" 'liHiil;fi,'J?::fl: l;..iti,*]
25
PSF pressure zofie.
-
Minimum direct soil load
18' (0.4t 13
(
26,667 PSI
26f67 X 0S67 X
s,
lJoncl
Weight of base
75 PSI
if S.
v- "12
x
(lB)3
B3
6
tl ro-
:
4M- .- 4x34'774 v'E * B-C', lB-8.41 --ru.+gotbs../ftr^:5.4inches . (1+12: 50,960 ) M, i'e' no compression KF:260X12X_li_steel
a;,O.i,|r$
d: 14 inches K': l,/B (C'plus d/6) : l/18 (8.42 plus 2.33) :0.6 D
C'plus C'e1
:34,774[b.tt/tt
100 up
18
3 82
-
and allowable bending
M.x12 x os6, d-:
:
fi-
as above and So
11
< 100
PSI
Use No. B @ 7 inches on centers
20,000
C':0.91 X 9.25: 8.42 feet B :
(2 gg
s.42)2ptrr-1
-
Wind Pressure: 0
and adding 33/s percent overstress for ACI 603 (c) f" : 1,500 f" : 26,667
o _M*rx6_
rs
17,256 plus 17,518
Concrete design for 2,500 Psi concrete @ 28 days f"
{
C')'z nlus
S.:0
M-
:99.9 14'
Uplift-Since fmin ) 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.
4Wo *o,:p _ 4 X 638,840 54 (18-5.64) in=2".y
:.r
soilpressure
/3 Sr, then let
X
0.867
854,540 lbs'
-
and obtai,
<
244,500 lbs'
1.801.200 -'---'---,
ande-
:
lbs'
x
x
Note: if S*
X
: *,, -
20,000 PSI stress So if Sw % Sr, compute
638,840 lbs.
18 610.040 1.Bo1,2oo 2
'Ertr
Consult
M"
M-
Design moment
: wo : Wr : W^ir: W*: Wt :
Minimum direct soil load
h'o:-:
438,420 lbs.
w"
-71)
Operating load on soil Operating liquid
156,265
(101 plus 28)
fill
Max. test load on
W-in Ww
: :
63,320 lbs. 610,820
*
B2
:
So
:
2,244 PSF
28,800 lbs.
582,020 lbs. 244,500 lbs.
soil Wt :826,520 *82: St:3,036
PSF
67
Seismic
factor 0.025 gives ?mio
M.:
651,960
( M.r
For example (1),
:##fl## :
:H
M" 4.33
)
1.5
610.820 X 16.5 ,lo:ffi:4.5-[ As rlo
)
Soil pressure (on
corner)
:_
16.5)3
1,482 PSF
: So * 1.414 Sw :2,2+4 + 2,096:4,340
5
Max. soil pressure
PSF
<
4,000 PSF
Concrete design for 2,500 psi concrete at 28 Days.
\o
)
3, procedure similar to example
(2).
Dowels. These must be provided to transfer any tensile force from the pedestal into the base. This force is transferred by bond from that part of the anchor bolts actually embedded in the pedestal. However it is usual to design dowels for the full tensile force whether the bolts go into the mat or not. Let N : number of dowels in a circle Dd inches in
Let M" : the wind or earthquake moment at bottom of pedestal Let A : area of each dowel; WR - minimum resisting weight
A:
W"
*
:
Wr)
45,100
X 2.25) :
483,520
(48lDd) M"
-wR SAXN
716,700
26,667x32
:
0.84 in2
faces.
: E#ii#
For
cases where
Mwr:
# **,
the depth of the base D is small
M"
compared to overall height of tower,
can be taken
as Mwf
Wo: (Wu * Wn) : 457,440 (+8/104) 1,801,200-457,440 _ A^_ 26,667xN -
373,890
26,667x32
:0.44
irP
i.e., provide 4 at No. 6 at each pedestal face For example (3), (+B/104 X 1,t09,580 6:ff:o'13inz ^_
457
,440
i.e., provide 2 at No. 4 at each pedestal face
Anchor Bolts. These may be computed as foliows:
: N: M': Wn:
Let Db
diameter
Then
*
(W"
For exampre (2), M"
:- 3,890 - 450 PSF
Subtract weight of original soil at given
depth 90,t
Wu:
+
i.e., provide 2 at No. 9 and 2 at No. B at each of B pedestal
M*rX6 _ l,log,58oxo Dw " _ ( B3
0.1 (438,420 X 56.75 2,600,470 Ib. ft.
)/.2,600,470-483,520 rr^ _(48/104) 26,667x32 -
3.6, Proceed as follows:
-
:
diameter of bolt circle in inches. number of bolts
wind moment at base plate minimum resisting weight
Then Root Area A
(48lD!) l{--
wR
fsxN
-
W* is usually the empty weight W", but when the earthquake moment is used, W* is the operating weight W" -F Wr, For example (1), Using SAE 4140 bolts with
allowable
stress 30,000 psi
Aboul the Author J. A. A. Cummins is a civil engineer working in design and construction for Hudson Engineering Corp., Houston. He attended Nautical College, Pangbourne, England, for four years-
M' :438,420 X 52.5
study civil engineering. He started his career with a consulting engineering firm in Scotland in 1947.
2,301,700 lb.
LJse
in
neer and superintendent on a project to construct. a darn in Scotland. He joined Hud,son 2/z years ago in Ontar,io,. and has been with the Houston ofiice for
a yeat. A
registered professional engineer, Cummins is a member of several technical societies.
M'
1
^4
i52
=-84)-lr*, :#X 1,801,200 - 1,d46,000 rb. ft. . (+B/10+) 1,646,000 - 409,620 \:@: 350,070
480,000-:
O.73 ir,2
Use 16 @ lt/a-inch dia SAE 4140 For example (3) M'
- SX
t,ror,rt0
:
1,013,930
(+B/104) r,01s,930 58,350 a:ffi:frpO6:0.41 - 409,620 Use 12 @ 1 inch dia. carbon steel. (/a" aia would do here, but normally
not be used)
68
_
For example (2),
England,
and from 1954-56 was an engi-
623,900
21-tt/s-inch dia. bolts SAE 4140
From 1951-54 he was'a concrete
structural engineer
ft.
. (+8/104) 2,301,700 -438,+20 _ A: @:600,000:1'u+1n"
then went to Royal College of Science and Technology, Glasgow, Scotland, for four years t(
:
inz
less than 1 inch would !! .tf
lf
NOTES
59
COMPUTER FOUNDATION., DESIGN e- E ;r \ * "d"- ""id_- -* 1
##e".
'% & L"
4 &-
"._
Btd
?
-:g"r*' j{Itr\* "1r ''
How to Calculate Foofing Soil Bearing by Computer Here's an effective method for finding the maximum soil bearing under eccentrically Ioaded rectangular footings, programed for a small computer
Eli Czerniok, The Fluor Corporation, Ltd., Los Angeles
Mosr on rrrE
srRUCTunBs used
in hydrocarbon
proc-
to some extent, affected by overturning forces which, iike the vertical loads, must ultimately be resisted at the ground. The function of their footings, then, is to provide that resistance; so that all the loads-vertical, IateraT, and overturning moments-can be adequately supported, without exceeding the safe bearing capacity of the soil. The factors and causes contributing to the overturning effects are 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 frlnway horizontal forces, unbalanced pull of cables, sliding of pipes over supports, thermal expansion (or partial restraint) of horizontal vessels and heat exchangers, reactiorx from anchors and directional guides, eccentric location of equipment are some of the additional reasons for the essing are,
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 in the various loading combinations met in design. The computations are the same rvhether the resulting overturning moment is from vertical loads which are located off-center (load times eccentricity); from lateral forces that are applied at a given height above the footing (force times distance to bottom of footing) ; or by some combination thereof. Therefore, the techniques for the computational analysis, described in this article used
FIGURE l-Computer-designed footings for a refinery.
will simply be based on the three resultants P, II, and M, for the vertical loads, horizontal forces and overturning moments, respectively; applied at the footing 6sn116id-'u/ithout giving any special consideration as to how this combination of forces and moments was obtained. It should be mentioned, however, that when proportioning footing sizes in the design engineering office, the actual make-up of the critical load-moment combinations could be of economic significance. Figure 1 shows several computer-designed footings in a refinery under construction. As with the other engineering materials, some increase is the allowable soil bearing is certainly justified when designing the footings for dead, live and operating loads, combined with the temporary lateral forces and moments. And due care must be exercised in establishing the proper design values. Obviously, no increase in allowable soil bearing would be advisable when the moments, about the footing centerlines, are due to the eccentricities of long-duration vertical loads. It should apply only to such loading
7l
CALCULATE FOOTING SOIL BEARING
..
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 ciose-up of a spread footing during construction is seen in Figure 2. IJnder 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 satisfactoriiy achieved, the engineer
FIGURE 2-Spread footing during construction.
combinations which are definitely known
to
include
overturning effects of a temporary nature. Building codes, recognizing the improbability of the absolute maximums occuring simultaneously, usually permit footings subjected to wind or earthquake combined with other loads, to be proportioned for soil pressures 33/3 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 Prqclice. 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 r.vell know, the distribution of loads and moments-on the footing, to the supporting earth beneath, is rather highly uncertain. Simplifying assumptions, however, come to the aid.
According to current structural engineering practice, the soil bearing under the loaded footing is calculated from static equilibrium, and on the basis of the simplifying assumption that the footing slab is absolutely rigid and it is freely supported on elasticaily 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 fuil area of the footing, and hence equal ,A to *. Wtl".r moment is also present, its contribution can be evaluated from the simple flexure formula p.o.,ided that the resultant eccentricity (.o--
$,
"
puted from $) frffr within the kern of the footing area. By superposition, the maximum and minimum pressures are simply the algebraic additions of the direct and bending components,* * Y u"d f - +, respecin pressures, increase the net obtain tively. In order to the weight (per unit area) of the displaced earth and backfill should be deducted from the gross values. In
72
must then tackle his next item on the agenda-the structural design of the footing itself. To accomplish that, he expediently relaxes the rigidity restriction, and permits the soil pressure against footing bottom to deflect upward (not too much though) the outer portions of the footing. To resist them, steel bars are added to compensate for the inherent tension deficiency of plain currcrvr". 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. Bioxicrt Eccenlricity. When the overturning moments are about two axes, the footing obviously, will bear most heavily on one corner, and least on the corner diagonaliy opposite. As long as the eccentricities from the resultant Ioad-moment combination are sufficiently small to remain within the kern, the entire footing is under compression, and corner pl'essure can be computed from the well known formula
PtM,a*tMrc, AIxI, Flowever, as the eccentricities increase and
fall
outside
computations become quite complex, even with the simptifying 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
the kern, the
that portion of the footing area over which the soil pressure would have been negative. It is the difficulty in deterrnining the shape and size of the remaining "effective" portion which constitutes the major stumbling blocks in the efforts to achieve a rnathematical solution. Depending on the location of the resultant of the applied
loads, the effective portions of rectangular footings could well vary from a triangle, through trapezoid, to a full rectangle. The line of zero pressure (neutral-axis) establishes the boundary of what is to be considered as the effective footing area. From statics, the value of the resultant of the applied loads P must equal the total
'lo I
J'lr __t D T
l
T .o
.o
x
l_o,
F
I
To
Ior
I
-1 FIGURE
l<-
on the load location, the efiective
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 knou,n or assumed position of the neutralaxis, the maximum soil pressure under the footing corner equals the resultant loacl P dividcd
o
ty
e
- I
A is the effective footing ar.ea; eo* and eo, are the first moments of area A about the x- and \:- axcsi a and b arc- the intcrcepts of the neutral-axis line on the x- and y-, axes, respectively. The origin of the rectangular coorclinates is taken at the footing corner where the soil pressure is maximum. Depending on the location of the resultant load P (in the quadrant of the footing rvith the corner as origin) the effective area can be one of five possible shapes. The ioad locations that correspond to these shapes (with matching
ff,
T
r,r'here
cross-hatch regions) are shown mapped
in Figure
3.
Srobilify. For the resultant load p to be within the kern, the sum of the eccentricity-ratios in the x- and y- directions must be equal to or less than one-sixth, .E*E"1 ,."., D= + f < 7. The footing is then fully uncler
c
I
i L_
area can be one of five possible shapes.
of the rectangle is The intensity of the
pressure, and hence the whole area
deemed effective
in the
analysis.
maximum pressure (at the corner) varies from an aver-
age pressure P whcn the load is located right at f the footing centroid (zero moment), to twice the average, rvhen the load is at the edge of the kern. As the sum of the eccentricity-ratios increasc to more than a sixth, part of the footing area becomes ineffective in the analysis: stability diminishes and the maximum soil pressure increases to rncre than trvice the average. Theoreticaliy, the maximum soil pressure rvould approach infinity, and the stability zero, rvhen the location of the resultant load P is placed along any of the footing sicies. Though the abutting power of the soil migl'rt ofler additional resistance to prevent actual overturning, its value is rather hard to ascertain. Common engincering 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 ."vithout a dependence upon iateral soil resistance; r,vith a value of 1.5 being the minimum recommended. The r,veight of earth superimposed over the footing should be included in the stabilitv calculations. Regarding the resistance to sliding,
73
,/i DIMENSIONS D, T TOTAT LOAD P ECCENTRICIflES Ex.
COMPUTE:
COMPUIE MAX.
SOII
kr = lxv-
BEARING:
GIVEN
Po: .lll A-;-
Ei
= ks: kr= ks= k": kz
P
Gtox
b
YpQov
lxv- )GQox lox- YpQox lov- )GQov Qor- Y"A Qov- )GA
COMPUTE LOAD COORDINATES:
xP: D/2 - Ex Y"=l12-Et
RECOMPUTE N.A. PARAMETERS:
o=fffift .D
COMPUIE INITIAL
k, k, - k3kc = [iE=-ETG
NEUIRAL-AX!S PARAMETERS:
o = 2Dl too% b = 2r (rrentruc
II
compurE
rHE
croiEiilic- pnciilinrs
I i
I or errtcttvE* AREA: I IL---A, Oo*Q"r lor,lor, l*tj
IGNORE P" (INSERT DASHES)
ADD ONE
TO
CYCTE
COUNIER
o
p,
IT = z
=
T D TI -; - El
PoLI
STORE THE N.A. PARAMETERS USED
IN COMPUTING IHE EFFECTIVE PROPERIIES:
o o .
see Figure
74
OLDo=o OLD
vol' 42, No' 8' 5 0{ ..Corcrete suppot Anallsis by computer," Hydraarbon Processing & Petroleum Refiner, FIGURE'[-Logic pattern for computer program'
b:
1963'
b
SCIL BEARING ANALYSIS OF RECIANGULAR FOOTING
Exomple I SIDE D
SIDE T
AREA OF
l5'0c0
10'500
157.500 s0.FT-
CYCLE AREA SGROSS l 157.500 r0c.000 2 157.500 100.000
sr..,
FOOTING
GIVEN LOAD
250.ooo KIps
H/ECCENTRICITIES
XY r.ooo
l..0oo FT.
PROPERTIES OF EFFECTIVE FOOTIN6 AREA
O
Q
ox
I
0Y
s26.875 g26.ttts
1181.250 lrBr.25o
APT.O lL EEARI|\G Ar ,coRNER 1129. t46 psF
Exomple
P
I
ox
5788.L25 5zBB.t25
APT.I
1317.370 pSF
I
oY
-START ITI TH ozor.ssz 620r.562
llst^2.500 ll8tz.50o
APT.2 50.066 psF
PARAI,IETER S
AB 30.000 2l.ooo 36.964 18. I t3 36.964 18. I t3
xY
APT.3 ts5l.84l
psF
2
D SIDE T 15.occ 10.500 SIDE
AREA CF
FCICTING EIVIru LOIO P
I57.500
sc.FT.
CYCLE AAEA IGROSS
I 157.500 lo0.00o 2 ll8'231 75.067 3 86.690 55.041 4 68'778 43.66s 5 6l'483 39.036 6 60.051 38. l3l 7 60.000 38.095 SOIL BEARII\iG
Krps ,]rro
,l,rro FT.
PROP[RTIES OF EFFECTIVE FOOTING AREA
C
0x 826.875 5ol-77e 298.22L 2a2.o49 166.s73 L60.272 160.0oo A PT.
AT
ro0.oo0
},/ECCENTRICITIES
oY llBl.25o 746.soL 4A6.?57 351.a55 309.317 300.345 30o.ooo
O
coRr{ER 5000.000
O
PsF
A PT. I
I
I
PSF
aPl.2
PSF
FIGURE 6-Computer printout of Examples I and
it would be provicled through the friction developed at the footing boitorn.
common practice is to assume that
Com-puter Progrom. The most diflicult part
of
the
problem (in both manual design and in formulating the procedures for sequential electronics computatiori) is determining the position of the neutral-a;is whicir is taken as the boundary line cf the effective footing area. The basic computer r.outine develol;ccl for solving liaxial eccentricity problems in reinforcecl-concrete, d.escribed in..a-previous article,l can also be used to solve footing soil bearing problems. That program was modified, sJ that title headings and data in the printed results rvould comply with the usual nomenclature applicable to foot_ ings. The formulas lor the neutral-axis and the effective section properties are the same as given in the previous article, and therefore will not be repeated here. For background and development o[ the formulas and criteria the reader is also referred to the writer,s paper ,,Analytical Approach to Biaxial Ecccntricity.,,: The logic pattern used in formulating this program for the srnall computer is shown flow-charted in Figure 4. From start to finish, loading of the program deck and data carcls, computations and
I
ox oY s7aB.L25 rrBl2.5oo zgss.846 6696.222 1516.686 4044.46' 88B.B8z 2788.659 680.168 2334.r52 64L.569 2253.Or8 640.OOO 2250.OOO
PARAIqET ER S
AB 30.000 2t.ooo 21.786 t2.3L7 L7.422 10.148 I 5.663 8.7e8 15.094 8.147 15.003 8.006 I 5.000 8.000 I 5.000 8.000
XY
START I.II TH
620L.562 2688.006 I 289.866 79 1 .004 630.
O
60
140
t.
75
600.000
aPT.3
.OOO PSF
2.
in Figure 5. Determine the maximum soil-bearing Ior a total vertical load of 250 kips (weight of concrete foundation and earth backfill included), located ecccntrically rvith respect to the footing centerlines, at a distance of l'-0" from each center]ine. shown
Solution.
: 250 kips D*: l'-0"
p
EY: I'-0" d
L
thc print-out of results for the trvo of a mintrtc.
examples cited, took less than one quarter
Exomple t. The plan of a footing used in a rigicl frame structure supporting several heat-exchangers is
FIGURE S-Example
1.
75
footing dimensions, see Figure 4), the correct location of the neutral-axis line is obtained within the first cycle. Note: The test for convergence requires that the neutralaxis parameters remain the same throughout two consecutive cycles, and whence the extra cycle shown in the computer printout (Figure 6) results. Having determined the position of the neutral-axis, the computer next calculates and prints the maximum soil bearing (at corner used as origin), as well as the bearing at the other corners. The soil bearing diagram shown in Figure 7 helps visualize the results.
CALCULATE FOOTING SOIL BEARING
)"::4'tgY!4-" -
Exomple 2. What is the maximum soil bearing, if of additional overturning effects on the structure, the load specified for example i is reduced by 150 kips of uplift, while the eccentricities are increased by 2t-9" and 2t-3", in the x- and y- directions, respectively. Draw separate diagrams of the soil bearing under the footing for both examples. because
.-H..,i'.*}*=
150 : 100 kips : 1'00 * 2'75:3'75 ft" EEccentricities I E,, : r.oo + 2.25 :2.2b ft. stability I s'R* : 7.50/3.75:2 0 Ratios! S.Rr: S.2S/2.25 - 1.62 > 1.5
Solution. Resultant P:
250
(
Pov
=
o'o+
*"
The load is now outside the kern, and with the large 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 eccentricities used
FIGURE 7-Distribution of soil bearing under footing.
For these eccentricities, the resultant load is obviously within the footing kern. Ilence, the full footing area is under bearing, and the position of the neutral-axis line, falling outside the footing, has no effect on the geometric properties used in subsequent calculations. And since the computer program was set up to start with the full rectangle (by using neutral-axis parameters equal to twice
About lhe Author Eli Czerniak is a principal design engineer with The Fluor Corp., Los A ngeles. He coordinates comPuter
ffi:;d
Engi ljesrgn [ngrfor the Design apphcatrons tor applications neering Dept., reviews manual tech-
;,,ffi"
;;";.;;i;;" ;;';;,h"d'
and procedures better adaptable. to systems converslon
rn
automatrng
the design and drafting of units.
in
refinerY
Mr. Czerniak received a B.S.
engineering from Columbia Uniin 1949 and an M.S. in Civil Engineering from Columbia in 1950. He is a registered engineer in Caiifornia and has published a number
versity
of
technicai articles.
He has had
field experience as a civil
engineer
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 Czerniak
his present position.
76
two rectangular components. These two component values are shown in the readers' interest. The overall stability' ratio for the diagonal direction was also computed, and found to equal 1.84. Now, with the resultant load being outside the kern, part of the footing area must therefore be neglected. Noting that the eccentricity in the, x-
.D
direction equals !; , it is apparent that the location of the load is on the dividing line between types I and III (see Figure 3). The limit of type I effective area is reached when parameter a becomes equal to dimension D, (and at which point t)?e III begins) . By observation, then, parameter a is known to be equal to 15.0 feet Such deduction would, of course, be helpful in reducing the volume of computations when attempting manual solutions. With a digital computer, however, the more generalized the approach, the better. The results are achieved by following the systematic procedure of successive substitutions of neutral-axis parameters to absolute convergence, which for this example was reached in six cycies (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 "Concrete Support Analysis by Computer," Hvonocemon No.8, ll7 (1963). Pnotr:ssIrc'ern -'iCz".oiuk" E.. Petnorruu Rrinea,42 "Analvtical Approach to Biaxial tccentricity." Journal of the Siiucrural Div., Pr6ccedingi'of the Amercan Society of Civil Eneineers. sr4 (1962), ST3 (1963). rCzerniak.
8..
concrete support Analysis by computer Axial Ioading plus two-directional bending in reinforced concrete supports is an easy problem for a srnall computer using this simplified program
Eli Czerniqk, The Fluor Corporation, Ltd., Los Angeles IIanB's A cENERALTZED TEcrrNreue together with all the formuias especially developed for the iystematic solution by-a diejtal computer of 6iaxial .....rtii"ity 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 crimmed into the comparatively little memory rpu"" oi the small computer with a core storage capacity of only 4,000 alpha_ merical characters. The program is completely general and can be used for sections with symmetiical as *.ll u,
non-s).rnmetric steel arrangements, multiple Iayers of steel, sections reinforced with more than one ba, ,ir", ,rr,rrrui
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 corner. Many constructional components of structures used in the. hydrocarbon-processing industry for supporting heat exchangers, accumulators, drumsr'.o*pr"rro.s, iipirrg, etc., are subjected to various combinations of axial ioaJs ald bending moments. Because precise analysis, except in the very-simple cases, was found to be raiher difficult, structural designers in the past had rationalized them_ selves into some rernarkable oversimplified assumptions that very conveniently blpassed the otherwise tedious solution. Such attitude of ,,ignore it and maybe it will go away" is both wasteful and dangerous. As a rule, fun"c_ tional and more economical, slender structures. built of higher-strength materials, are now used in redneries to support much heavier and larger processing equipment than the massive rvall supportiof iays pastl Singie col_
umn tee-supports and rigid frames, such as seen under construction in Figure 1_, when subjected to lateral loading (e.g.,. from wind, earthquake, impact or vibration) ii addition to the equipment weights, often involve the load_ moment configurations requiring a stress analysis for axial Ioad combined with two-directional bending. The Neutrol Axis
in Reinforced Goncrete. The major manual _designs and in formulating procedures for- sequential electronic computation is th! determining of the position of the neutral axis, the in_ problem,
in both
FIGURE l-Single column tee-supports and rigid frames" termediary that is needed before achieving the final re_ sults. This computational complexity in reinforced_con_ crete stems essentiaily from the cornmon assumption that part of the section is considered ineffective for design purposes (cracked-section design) . Thus, even when the shape of the cross-section of the reinforced-concrete mem_ ber might be a simple rectangle, the shape of the con_ crete's_ effective portion (used in analysis) need not neces_ sarily. be one. Depending on the relative values of applied bending moments to concentric loads, the shape tn. "fvery concrete section to be inciuded in the analysis could well vary from a triangle or trapezoid to a full rectangle.
The fact that the effectiveness tf tt e reinforcing steej is not always considered constant tends further tJ compli_ cate the anaiysis.
Slress in Concrefe. In reinforced concrete design, the concrete itself is generally not relied upon to withstand much tensile stress. (The reinforced-concrete as a who,le though is .quite capable of resisting significant amounts of eccentric tension loads as will be ,io*., in Example 2.) It is usually assumed that the tension stresses in ihe flexural computation are taken by the reinforcing steel, whereas the compression is primarily resisted by tie con_ crete. According to Section 1109 (b) of the AbI Codex * Building Code Requirements for Reinforced Concrete (ACI 3lg-56)
77
CONCRETE SUPPORT ANALYSIS
.
.
I
'ttx.v = 9]-3
FIGURE 3-The area under compression is a triangle' FIGURE 2-In nates in oroe corner.
rectangular sections, locate origin of coordi-
some tension stress
in the concrete is permitted when, in
stresses, there also exists direct compression and the ratios of eccentricity to depth (e/t) is not greater than /3 in either direction. Assuming a straight line stress distribution the stress at any point (x, y) in the concrete may be written:
addition to bending
fo,,y
: t" l- _; * _ i_l "l L,
where fo represents the intensity of stress at the chosen point of origin, and the constants a, b designate the interiepts of the neutral-axis line on the x- and y-axes 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"*, , would be negative is said to have thus become ineffective for purposes of analysis. It is apparent from the stress equation that the region over which fo",, is negative extends to all points for which the value
i * * is hrger
It is evident, therefore, that in cracked sections the intercepts a and b can be also used to denote the boundary line of the concrete's effective section. Convergence of the two lines until they almost coincide constitutes, 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. The area of this transformed concrete is assumed to be concentrated at a point which coincides with the center of the replaced bar. The amount of concrete resulting from the exchange depends on the relative effectivenes attributed to the materials. In the strictly elastic analysis, the modular ratio n is the index to measuring the relative effectiveness of the steel over that of concrete. The area of the concrete substituted for each bar equals n times Ai. Of course, it presupposes that the bond between all tension and compression bars and concrete remains intact at all times, and they deform together under stress. fn reality, this is not exactly true. There is experimental evidence that the bars in the comthan one.
pression 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 cornpressive reinforcement. The allow78
able stress values are well above that which might have resulted from a strictly elastic analysis. Section 706 (b) of the ACI Building Code requires that: "To approximate the effect of creep, the stress in compression reinforcement resisting bending may be taken at twice the
value indicated
by using the straight-line relation
be-
tween stress and strain, and the modular ratio n." IIowever, 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 f1 the equations governing the stresses in the reinforcing steel can be written as:
tensile f courpressive
":
f'.:
*"
I-
*
"l L'-; - -il1
[ " "l 2nf,fr-;ol=,,
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 co,mpression 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:
l- " "l f"l1----;-lA, "L bl u
'
With the transformed area concept, the correction
is
accomplished by reducing the effectiveness index m by one. The area of concrete which is substituted for a bar in compression would be equal to [2n- 1] or less, times Ai. 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. Ifowever, in the limited cases when tension in the concrete is permitted, these bars also displace some eflective concrete. Hence, they too must have their areas subtracted or the
lOOo/o
Compression
FIGURE 4-Variation of five modular ratio modified by using
Copocity
("_ l)
instead
of
n.
of Looded Section. The magnitude of the
largest load which can be sustained ut .-gir.r, location (within the prescribed stress or strain iir"i"l constitutes the measure for the capacity of the scction. F"; ;"t known .or assumed position of the ne,rtral_a*is it can b! determined with ease from the equation as follows: Eccentric Load
p: t ["- + - +]
Where A denotes the over-all effective area of the cross_ section_and Qo", Qo" are the firrt *o*errts of this area about the x- and y-axis, respectively. In most- practical . .problerns, howcver, the position of the neutral-axis is neither known .ro. .un it U" ."u.orruUif assumed. Given data usually include the magnitude ani the position of the imposed load, as rvell as the material
specifications. The problem then becomes one of deter_ mining the adequacy of the section to sustain , gi.r"rr-a"_
sign load, u",trrg at a given point, and not exJeeding a given stress limitation. -The tcatitn of the neutral_axis may be_,-in itselt of very little interest to practicing engi_ neers. Nevertheless, it must be determined first, "U"fo'r" proceeding with the more essential task of structural adequacy. The general equation* fo. "rtut[rfrfurg tfr" fui
rameters
of the neutral-axis
-
(L.
-
ypgy) (r."
Where I"*
x,
an_d
y-axes, and
- xo&.)_-
rectargle.
obtained. Furthermore, by choosing (as the origin) that corner at which the concrete .o*pi.riirr" stress is a maxi_ mum, the number of possible shapes of effective concrete
area is reduced to five.
In Figure 3 the corners of the given section are O, B, ^ and D. Line C, QR designates th"e ,r",ri.ui_u*i., and'in_ tersects the-x- and y-axis at a and b, respectiveiy. When tension in the concrete is not permitied, ihe neutral-axis line is also taken to represenf the boundary line of the portion of the concrete section considered effective in the analysis. When the neutral-axis intercepts are smaller ,F;.orresqo.nding dimensio", of section (as ",1:" tn..t'.igur,e 3) the area of concrete ifr" ,nder comprls_ llyl slon ls a tnangle. As one or both of the intercepts are increased beyond the section's dimensions, the effective urea p.ogr".r", from that of a trapezoid to one of a rectartele. When the neutral-axis falls completely outside the section, the whole area is obviouslyunder compression and therefore fully effective. The variation .f th; h;;;'af,es, from t.i_ angle to rectangle, are shown shaded in Figure 4.
a:
IYeQo,) (roy-xpeo") ro*
y_axis intercept
(r,y
to
are:
x_axis intercept
(r*r-Yrq"; (I-"-{&l
V shapes frorn triangle
(rs,
b:
_ yoQo*)
-
(roy
yreor)
_ xeQoy)
are the. moments of inertia about the
_Io, I," denotes the product of inertia of the
area about the origin. Xn and yo are the coordinates of tne apptred eccentric load. In the above equations, all the section properties obviously pertain to the over-all effective section.'Th" p.op"._ ti9s. o{ the effective portion of the concrete are added wrth the transformed properties of the steel.
Recfongulor Sections. In the case of rectangular sec_ tions, it is convenient to locate the origin of ,fr" coordi_ nate system in one of the corners of the rectangle (see Figure 2) and let the axes coincide wittr two sides. The main advantage is the relative ease with *hi"h th" ur.i_ ous formulas for the required section p.op".ties can be
*t':-xm**:+$+i*P;i,:,:g:*+,$H,B$'J;lf ;fl *ti'E*#
The required
properties of the effective portion of the concrete for the five possible shapes can be tbtained from
tne lormulas tor shape JV. The geometric properties in terms of the neutral-axis intercep-ts and the section di_ mensions are shown formulated below:
* * [,_(..),_ (+). ] q".:f "u,[,-(..)'-
AREA:
f =)--, (?'+] q",: f "r [, - (';')"- (r-)i-, (3 l)"+] r"":f .u,[,-(*J'* (';,)-_, (5),+(r.,t )]
:
f
;,)'* (+)'_, (+)" +("-J] _ ,.,: * "" [, (:_;,)-_ (o;,).n (+), + _, (,;.),+] r*
a.u
[,
-
("
79
CONCRETE SUPPORT ANALYSIS. .
N
.
Moments of Inertia
\t Io": lr
m Ai yi2
i=1
old poromde6 A ond
r^oy-I
N
Ai xi2 Ll m i=l
N
I
Product of Inertia I*":
I
mArxiyi
i=1
Convergence Technique. In the usual design problem, is necessary to find the size of the reinforced-concrete section which can adequately sustain a given system of loading. Several loading combinations must frequently
it
tiial sections must be incremented until all conditions are satisfied. The most laborious part of the computations (as design engineers well know) is determining the parameters of the neutral-axis line' The coordinates- of the applied load are calculated from the bending moments (usually given with respect to the centerlineJof the section). Together with the properties o{ the assumed section, they are used to determine the neutral-axis parameters. There will be only one neutralaxis which will satisfy equilibrium conditions and stressbe con-sidered, and the
Patomailt A
io axls.
-iest
L
strain limitations. When the concrete is not permitted to take any calculated tension, the neutral-axis line is assumed to be the boundary line of the effective portion of the concrete. The problem, then, is to find that neutralaxis which almost ioincides with the edge of the effective
FIGURE S-Computer sequence for properties of efiective 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 ratio n, 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'
section. The difference 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 successive substitutions until the desired results are achieved. To begin with, the distances of the load from the coordinate axes are determined. With them and the properties for 100 percent compression (with neutral-axis parameters equal to twice the section dimensions) the
of
first trial line is determined.
''
('-; - +)
About the Author
develops new methods an-d procedures better adaptable.to
systems conversion in automatmg the design and drafting of refilery ,nits. M1. Czerniak received a B.S' in ensineerine {rom Columbia Unirersiri in 194-9 and an M.S. in Civil Ensinlerins from Columbia in 1950. He'is a r"?iste.ed engineer in California and-has published a number
- r]= z"-r
where fo is the concrete stress at the origin.
Therefore, the required transformed properties for
of
N
A: l-
E 'Ai I
N
N
80
Qor:
T mAiYr i=1
Qoy:
T l) mAt x, i=1
in
engineer
design and drafting r,vith Arthur G. McKee Co. in Union, N. T.. for two vears before ioining Fluor in 1953 as a stru"ctural designer. He soon headed up the. structural design and drifting on various projects until assuming his present position. Czerniak
Morpent Areas
technical articles. He has had
field experience as a civil
steel reinforcement are: Area
the neutral-axis line falls
within the section, it is subsequently used to define the effective section, all the properties are recalculated, and
1 when For compressive reinforcement, its value is 2n comin the the stress When the bar stress is less than ft. stress) pressive bar reaches (the allowable steel tensile the value of m is reduced to
ft
If
and woiked
new parameters are determined. The process of substitut_ ing the calculated parameters of the neutral-axis for the parameters of the effective section is repeated to any desired degree of approximation. The routine
8or
"onrr".g"r"e is quite fast, and only a small number of cycles will
9,OOO' il,625'
usually be sufficient for most practical problems.
r5.625"
Compuier Progrom. When setting up a computer program for solution of engineering 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 tasis 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 appliiations. The program was written in SPS (Symbolic programing System) and punched into 529 cards, which were latei condensed into a 104 card deck. The card reader has a rated speed of 800 cards per minute, which means that it takes appro.ximately 8 second to load the whole program. Corn_ putations and printout average one-half seiond per cycle. Absolute convergence, wherein the neutral-axis paiameters (measured to three significant figures to thl right of the decimal point) remain the same through two cinsecutive cycles is usually achieved within eight cycles. In most instances the results of the third iteratiron seemed to have sufficed for all practical purposes. In the two examples cited, six .y.Lr *"." ieq"i.ea for the absolute solution. From start to finish, Ioading of the program and data cards, computations and printout of results-for both examples, took 15 seconds. Now, after the input data has been entered and machine digested, the convergence routine starts with neutral-axis parameters equal to twice the section di_ mensions and computes the necessary ,transformed, section properties from which, together with the load coordinates_new_parameters are calculated and subsequently used. Ifow formulas for the section properties foi only case IV are used in the program to determine all possible effective-section properties is illustrated in the blo-ck diagram, shown in Figure 5. A, .?"! iteration cycle the equilibrium load compati_ .. ble with the section properties, load coordinates, limiting stresses and the newly determined neutral-axis paral meters, is computed and printed together with maximum O 18.000 SIDE
SIDE
T
l4.ooo AR EA
COORDINAIES - 9.000 - 7.000 LOAD
xYs# ooI 0x
FIGURE &-Corner column in exchanger structure used in examples.
compressive stresses in concrete (at corner used for origin) and steel, as well as the maximum stress in the tensile reinforcement.
Numericq! Exomples. The reinforced-concrete section 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 shown
tteel.
Example 1. The column section shown in Figure 6 is 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 arrd 2r, I Ioaded
or
19
A
HEIGHT
2.640
t35o.oooPst
O
46.6624
760L
LOAD
OS
P ARAIT{E T ER S
XY
3160.0798 15876.0000 I 90 36 .
,2r20t
282.6450 244.8000 52f.4450
I,IODULAR RATI
NH 1,0.0 19.0
I OY
5530.6624 272 16.0000 327
SUII
9.012 14. 116 I
0Y
203.6100 208.0800 4l 1.6900
-
for compressive reinforcement.
lS.l
NO. OF BARS
06
2.375"
t5.625" il ,625"
ox SIEEL 50. 1600 351.1200 +51.4400 3530.7976 cNcRT 252.OOOO 1764.0000 2258.OOOO l6464.OOOO IOraL 302. 1600 21r5.1200 27L9.4400 I9994.?916 CYCLE I CNCRT FO I35O.OOOPSI STEEL,COHPR I9548 IENSION
STEEL cNcRT 6l.20oo '4.7200 IOTAL 95.5200 cYcLE 6 C|{CRT FO
Coordinoles
XY 2,375" 2.375" 2.375" il . 625" 9.ooo" 2.zts"
0798
1902.9860 325r.slL7 1770. t8l7 I061.2080 1468.8000 6;(.2400 2964.1940 +722.77t7 2194.42L7 STEEL,COt{pR l54l? TENSION le45{ LOAO t5r28t
AB 36.000
28.000
19.200
l5.486NFh
12.000
10.200
l2.oo0
lo.2ooNE,,r
FIGURE 7-Computer solution to problem Example
8t
CONCRETE SUPPORT ANALYSIS. .
:
Solution. With P
15 kips, the eccentricities are:
22'5 x €*: -
12
:
x
12
:Il
l5
ey:-
is an actual comPuter printout. For these eccentricities the load coordinates listed with resPect to one corner as origin are:
.
t7.5
15
inches XP: -9 inches YP: -7 After convergence, the load caPacity is shown as
18 inches rnches
15,128 ibs. which is slightly more than the given 15 kips and hence O.K. The capacity of the section to sustain this eccentric load was evidently limited by the 1,350 psi compression in the concrete. Maximum tensile stress was
The solution of this problem is shown in Figure 7, which
14.64
Y
K
lg,4i4 psi which is close to the limiting 20,000 value given; maximum compressive stress in the steel is 15,417 came out to 12 isi. The final neutral-axis parameterswhich means the inches for A and 10.2 inches for B, shape of the effective portion of the concrete section was a triangle (case I in Figure 4) .
(tension)
Example 2. What is the maximum tension load which if the eccen-
can be supported by the section of Figure 6, tricities of Example 1 are halved?
: ? With eccentricities equal to e' : 9 inches and ey: 7 inches it is apparent that the load is Solution. P
Iocated at the corner of the section, and since must be opposite the corner used as origin.
SIDE
D
SIDE
18.000
XY I 8.000
14.000
A Sf 2.640
OF EARS
LOAD COORDINATES
T
06
14.000
sUM
IIEIGHT
O
I{ODULAR RATI
14.116
9.0 I 2
it
is tension,
OS
NH 10.0 19.0
I
ox
27.8109
CNCRT TOTAL CYCLE
58.1709 cr{cR
T
t880.6491 231.0659 zltr,1L50
194.2050 241.OO50 12.7L62 65.7356 259.9406 319.1212 760.456P5I STEEL'cot',tPR
30. 3600
STEEL
FO
FIGURE
1685.r)67
2971.21L7 285. l9l0
128.9077 1814.4444
)2L8.4047 5536 TENSION 2OOOO LOAD -lq676d
&-
7.844
7.09 I
7.9ttq
7.09lNf
h
Computer solution to Example 2.
The computed tension load capacity TRANSFoRxtO PRLPERTIFS 0F STETL RFINFORCING PARAH€TT( B
ITR A l2.oo0
PARAHE
lxYXOo. s .\40 t5.625 11.625 10.0 .440 2.275 2.175 19.O .440 2.175 t1.625 tO.0 9.000 2.175 l9.o .{{o .4(O 9.000 It.625 to.o .440 15.525 2.175 lO.O IIIIGHT SUH T0TALS 9.0121 l4.l16
O
1(.1200
7
A
X
Y
T0IALS
IE I GHT
51.1500 10.4500
oY lO14.2l87 \1.1556 24.8187 677.1600 156.4000 lo7q.2ln1
203.6100
252-6450
I902.9860
r7.tl.r?17
UX
t9.8550
51.1500
I9.8550
B
N AIiO I t9'o lo.0
XOOUI,AR RATIOS
?.09t
llo0.
s
SUX
O
9.Ol2r t4. I ]6
06
PRoptRIltS 0F SThEL RFINFURCING 8^RS (fOr eXOmple 2
.A44
.\40 15.625 11.625 10.0 .440 2.375 2.f75 l9.O .440 2.375 11.625 l0.O .440 9.O00 2.175 lo.0 .440 9.000 11.625 L0.o .440 rr.625 2.r75 10.0
NUHETR OF 8AR5
0x 594.6187 41.1556 594.6181 47,1556 5e4.6l1l 24.nlal
A
51.1500
PARATETER
A
RAIIOs N ANO N lrl.0 lo.0
atll oY ha.lrca 19.8550 10.4500 15.?4OO 39.6000 68.7500
ARIA 4.4000 a.1600 4.(000 8.1600 4.4000 4.4000
TRANSF0RHTO
PARAHETTR
HOOIiLAR
10.200
sARS (fOr exOmple l)
o UX 51.1500 19.8550 5t.1500 I0.(500 51.1500 10.4500
68.7500 19.8550 1o.1500 39.6000 19.6000 68' 7500
lrI rix 5s4.61A1 1f.t556 594.6187 24.818I ,14.61A7 24.8187
30.',,600
194.2050
247.0050
1880.&4',1
o OY
l2l.\ALZ
expected is way below the allowables'
l6l.2Al2
example is to illustrate the often overiooked 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 sup-
I78.6?50 {60. l50o
l770.l8l
7
)
06
0Y li)74.2I87 (7.1556 24.818I 156.4000 1t6'qAOC lo1c.2lb1 297','2llf
FIGURE 9-Properties of reinforcing bars for Examples
82
47,1556
NUHETR OF SARS
4.4000 a.15oo 4.{OOO 4.4000 4.4000 4.4000
lRfa
xY
799.2lb1
1
and 2'
is 14,636 lbs., limited by tensile reinforcing stress of 20,000 psi. (Figure B) Maximum corner stress in the concrete (at the origin) is 760.456 psi and maximum stress in comPressive reinforcement is 5,536 Psi, which as
xY
799.2187 47.1556
12I.4812 94.0500
4a,0.1500
161.28t2
1685'51'r7
The main purpose of the
second
plementary routine was written that prints out all pertinent properties of ihe reinforcement for the converged parameters. The comPuter Printed properties of the steel reinforcing bars for both Examples I and 2 are shown
in Fisrrr'c 9.
##
NOTES
83
FOUNDATIONS
t( solrF
k_1
g
{
I €
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 Mqkqretz, The Badger Co., fnc. Boston, Mass.
Tooav's pETRocrrEMrcer, plants are being constructed locations and under conditions that require more attention to foundation design than was in the ".,rto-a.y past. New plants are often close to water, on filled sites, where the land is not ideal for foundations. In .""..ri years the trend has been to higher towers, often com_ bined in groups; equipment has become heavier. Moreover, rigid rcinforced concrete structures permit only negligible differential settlement. Tank foundations de_ serve particular attention.
in
New Design Techniques. These considerations suggest the desirability of design innovations or nonconventional
design techniques. Foundations having negligible settle_ me1! 9a1 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 settiement of fills, placed over uniform deposits- of clay, than it is to p."di"t deflections of pile foundations ioaded by a structure and subject to down_ drag load from subsiding fills. An error in ihe 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 indled. Foundations on sandy soil will settle quickly and will be stabilized, pro_ vided no considerable change in subsoil water level occurs. Foundations on clay settle slowly and over a longer period of time, the settlement aiso depending upon water level variation, but not to such an extent as it does in the case of sandy soils. Storqge tqnk foundqtions appear to be unimportant structures in petrochemical plants. Ifowever, considering the large investment in tanks, substantial economy can be reali"ed if, by proper founclation design, long main_ tenance-free tank service life is achie.,"J. L 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 .nave, aiter 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 5O feet high there will_be approximately 130 psi under the shell and 23 psi in the *lddl" of the iank. A large differential settlement between the shell and .bottom may cause a tearing or shearing effect between the bottom plate and the she'il. Ho*"r"r,'lu.ge tanks over i50 feet in diameter can be used if aim".."iid settlement w-ill
85
Tonk. f
= 50' H =30'
2'
,%Z: '3'(Port
,/,
v%)
Cloy t2'
Z
o)
//zzztzlb)
CtoyiT= l15 lbs. per cu. ft. C= 8OO lbs. per sq.ft
6'(Port
FIGURE l-ExamPle figure.
is as large as 24 inches, because of the flexibility of the bottom ind the roof plates. The effect of the relative settlement between thi tank and the connecting pipes can be overcome by using flexible joints. Differential
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 /2 inc]n around
the perim-
eter is necessary for proper tank erection'
Even small locaiized deflection of the foundation during operation may cause "hang-up" of the floating roof.
.
tt is eosier to Predict the settlement ol tills over claY lhqn
Edge cutting under the tank shell may cause rupture of ihe weld between the tank bottom and the tank
looded pile dellections subiect to downdrog from subsiding fills
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.
settlement for small tanks (up to about 30 feet in diameter) should not exceed abott t/z inch. If the dif-
ferential settlements under the shell itself are closely spaced, e>
.
IJse
a sand cushion as above with
edge protection
consisting of a crushed rock ring wall. o l]se 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
con-
troversial subject. Some owners feel that "edge cutting" is not detrimental and that the cost of the edge treatment is, therefore, prohibitive. Others are of the opinion
86
.
Some tanks need anchorage (aluminum tanks or tall
tanks having small diameters).
About the Author Iohn Makaretz is the chief structural engineer with Th"e Badger Co., Inc., Boston, Mass. He has had a wide .experience in slructural, design .i;.,l.l.;li* in building dams, bridges and heavy .ii industrial lonstruction. After receiving an M.S. degree in engineering
from Lwow Institute o[ Technology in Poland, he practiced structural engineering in Europe for several years. Before joining Badger, he was
chief structural engineer with
f,,Tm#r"iJ"; f,:-il:. l?'if;
i! i# ''l l.$}$
rrll }i;1 fi;:::
il
American Society of Civil Engineers, :r1'.;"r @llS,i Association for i;;: : :: :;i!i:*i';n
the International Bridge and Structural Engineering Makaretz and the American Concreie Insti-
tute. He is a registered professional engineer in the State
of New York, New
Jersey and several other
states.
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 approximaiely 20
feet thick, the following methods of foundation design can be used:
o If the thickness of weak deposits is relatively shallow (3 fo 5 feet), it is often advisable to remove the weak materiais and replace them with well_com_ pacted granular fills. Note that it is necessarv to extend the compacted fill beyond the tank pe.i*et"r.
. Io. deeper, weak soil deposits, it is entirely prac_ tical to surcharge the compressible strata before the tank foundation is constructed, if time permits. The purpose of such a surcharge is to increase the strength of the subsoil and to reduce the tank settle_ ment during operation.
1.5
C
5 o,EO''' o
.9E oo
tz
$ F rr S s'u o,^ o0s (/)5 o8
ole
EE-
O+O6
oo E*ou
o The tank foundation may be put on a crust of very strong filI and allowed to float on weak soil strata. This is practical where the ground has to be filled ?nyway. The crust must be thick enough and extend far enoLgh.beyond the tank perimet-er to prevent lateral plastic flow of the weak subsoils. Ste;l sheet pilings, concrete rings, or crrshed stone rings may be appiied to prevent lateral flow of the weak sub_ soils which might cause tank foundation failure.
Sheef sfee/ pilings, concrete rings, or crushed stone rings mqy be used Io prevent lqteral flow ol weqk subsoils
25
pressed as:
gurt: C X N" * y X a
(assuming that the clay is saiurated, an angle of shearing resistance /: 0)
The cohesion of th^e soil (C) for our purpose may be assumed equal to 50 percent of the nonctnfined com_ pressive strength of the,soil. The bearing capacity factor N" varies from 5.2 for elongated footincs"to 6.2 for round and square footings. For rectangular footings, N" :
0.84
+
0.16
f
X N" (for
Ol The Totol plont ln Million Dollors
FIGURE 2-Soil investigation cost as a percent of founda-
tion, structures and buildings
Where
cost.
B is the width and L is the length of
should be checl
to Skempton's formula, N" : coefficient 6.2 for round and footings (nondimensional (adjusted t
eurt
:
5.44
X
: Liquid load : Pad weight :
-- 5.2 800
Tank load
+
+ -
5.44
X
:
0.75
120
1442lbs.per sq. ft.
1900 lbs. per sq. ft. 240 lbs. per sq. ft. r,rr. per sq.
. +442 saletv: ' 2205 =
2) Using the balancins 9,r,X
for strip
65 lbs. per sq. ft.
Total: ZZ* -Factor ot
F
square footings, 5.2
)
ft.
2.
moment method:
2XCXL 'I
-:CYLXT-5'44 Reduction factorp. 6.20
=0.88
2\B00xi2x3.1l _: 5024 1bs. per sq. ft. ----12Reduced es11 : 5024 X 0.BB :4421 lbs. per sq. ft. qurt
--
(the result should be the same as in Case 7,
square or circular footings)
the
rectangular foundation, in feet. Although it is very important to establish the ultimate bearing capacity of clays in which shear failure may o:c-ur mgre frequently than in noncohesive soils, never_ theless the settlement probability for the foundations on clay should be considered and its expected magnitude
N.
Stability analysis of cohesive soils may be made using . either A. W. Skempton,s methodl o. the balancin[ moments method belween the imposed load and the shearing stress resistance of the soil strata in question. In order to calculate the stability of the tank founda_ tion, properties of a clay stratum are required, such as: undrained shear value (C) lbs./sq. fi., aensity 1y; lbs./cu. fr-., bearing capacity factor' (nondi-".rr;o.rri)', F_" and the height of the surcharge (d) ft. According io Skempton, ultimate bearing cipacit'5, of clays ir'""_
o/"
Cost
per sq. ft.)
Settlement
or
4442 lbs.
for above conditions. Assumptions: L,,, 87
A
pile toundqlion for tsnks with
q
reinforced concrele slob copping is besf but the most expensive; or, o -foot capping ol crushed sfone compocted between the piles trqnsters tonk lood to piles well
liquid limit of clay : 40 Percent; eo, void ratio : L'2; factor : 0,009 (L;'o) : O'27 ' Co "o*pr"tsibility It is assumed that this strain is constant for the clay stratum and the clay is normally consolidated (i'e' no drF.g 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 investigotions ore c, smoll pqrt
of {,otol plont cosls Yet
some owners obiect to toking o sulficient number of borings or ony at qll
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' Ilowever, the necessity of releveling the tanks several times during installation still may save money as compared with pile foundation construction. The conveniional pile foundation for tanks, with reinforced concrete slab capping, is the best, but the most -expensive. in i"itf, tn" tank cost. Alternatively, ? -caPPing "o*pt.i*r, about 4 feet deep, compacted between stone, of ciushed the piles, may be used. The compacted crushed stone arches between the piles and transfers tank load "."rio to piles relatively uniformlY' 'i-he exact prediction of tank settlement is impossible, except if supported on point-bearing piles, for the following reasons: o The stress distribution in thin, weak soil layers under the foundation cannot be accurately determined' o The magnitude of lateral plastic flow in highly stressed soils is unknown.
of the crushed stone caP on the piles difficult to predict. The nature of the deflection of piles in soft soils
Behavior settlement calculation the 12-foot clay stratum is divided into two 6-foot layers. Approximate settlement (in.),
A: H X
unpredictable.
12
y - l- ro*,n p*ap Ite" For part a, p:2' X 120 : 240 lbs. per sq. ft. 3' X 15 : 345 lbs. per sq. ft.
Relcrtive Cost of Soil lnvesiigotion. Let us consider an average size petrochemical plant, the total -cost of which
is abJut six million dollari. The approximate cost of foundations, structures and buildings would be about 25 percent of the total cost, or $l/2 million. Soil investi-
1
3a;Ibr. per
ap:
1900 lbs. per sq. ft.
f
sq.
65 lbs. per sq.
ft.
ft.:
gation for such a plant would require about 10 borings,
1965 lbs. per
sq. ft.
: 585 * 1965 : 2550 lbs. per sq. ft. _ 6'x 12 Y 0.27x logro 2550 2.2 ^" -85-:5'l For part b, p - 2 X 120 : 240 lbs. per sq. ft. 9 X 115 : 1035 lbs. per sq. ft.
Plotting soil investigation cost against 25 percent o{ the total plant cost (foundation, structures and buildings) we obtain a curve shown in Figure 2.
rn'
tu.. per sq. ft.
*Ap : 1275 + 1765: 3040 lbs. per sq. ft' (Ap for the part b, decreased in accordance with
Boussines q
formula) 3040 12 .', lo1ro *W5 :3-4in. Z2 ,.0.27 X
6',Y
Totaldeflectio"=gi*
It is assumed for the above investigation that 88
The cost of soil investigation is small if it is related foundations, stmctures 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 arry at all. A comprehensive soil report an engineer to design with confidence, repays the "rriblo the soil investigation, and saves money for the
to
p
^-a-_
laboratory analysis and a complete report would amount to from $4,000 to $9,000, depending on
*hi.h *ith
soil conditions.
p -l- ap
lzzs
ls
:T:":l
LITERATURE CITED
A. W., "The Bearing Capacity of Clavs," Building Reseach l95l. Consress. -;Fim.tairer of the American Society of Qivil Engineers, Tte Journal of r Skempton,
the tank
Soit Mechanic"s and Foundatioro Div., Part 1, Oct. 1961.
Graphs Speed Spread Footing Design When designing square or octagonal footings, these graphs will cut the calculation time to a minimum. No trial and error sizing is required
F._ B. vqn Homme, Chief Structural Engineer, Fluor-Schuytvlot N. V., Haarlem, Holland*
Tnrar,-eNo-rRRoR srzrNG of spread footings can be supplanted by a better method. A graph can be used to size footings with a .ni.ri-.,- of calculations. IJse Figure 1, to determine the size of square footings, Figure 2 for
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 Iateral 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. Ffowever, 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 are divided by the soil pressure coefficient N.
N:
octagonal footings.
Yl000:
Stability and soil bearing are the main considerations 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
M1000
in
* Subsidiary of The Fluor Corp., Ltd., Los
Angeles, Calif
:
allowable soil pressure in lbs. per sq. ft. 1000 lbs. per sq. ft.
total vertical load in kips N overturning moment in ft.-kips
N
Squore Fooiings. When the footing can turn on the A-A-axis only (Figure 1) (pipe-rack footings, for ex-
t2
ro
a.
a (L
4
:<
t
I
3.5'
?4"4r, 'aq
A
kP3.
+'R
D=5'
)=5'
o o o
5
R
Vlooo
to
25
NOTES'
(KIPS)
l. Dofted D'lines oreto be used for stobility conditions ond the solid D-lines for soil beoring conditions when footing con turn os well on A-A-oxis osB-B 2. whenfooting conturn onA-A-oxis only the
dotted D-lines ore to be stobility os well os soil beori ng condit ions.
-
FIGURE
l-Use
used fur
this graph to design square footings.
89
{i
bo
)E )E uk
h0
(,
o b, a c)
Je
o
* !
EO
o 9
a
D I
N
ae (Jri
lr* .H I
I
I I I I
I I I
oA ro
o)<
o o o
t +-
90
(Saty '131
OOOI
Y1
ample), the dotted D-lines are to be used for determining
the size of the footing. When the footing can turn on the B-B-axis as well as on the A_A-axis, the dotted D_ lines are to be used when stability govlrns. But when soil pressure governs, the solid D-lines ire to be used. The stability is smaller when the footing turns on the A_A_axis than when it turns on the B-B-axis. On the other hanJ, the soil pressure is greater when the footing turns on the
SQUARE SHAPED Welp,ht (klos oer wrdth (ft.) l'of Hetght) 2.5. .
OCTAGONAL SHAPEI) WeiEht (klDs rrs 1" of Hetght)
.. . .. .. ...
3.0............ 3.5............ 4.0............ 4.5............ 5.0............
B-B-axis.
0.&39 1.036 1.250 1.490 1.750 2.O20 2.330
Ocfogonol Footings. In Figure 2, the octagon was replaced by the inscribed circlJof the octagon. Th. diff..ence in_ soil p:.essure between an octagoial footing and a circular footing under the same loa"ding conditilons is not important.
2.650 2.990 3.350
3.740
4.tN
Y1000
When and M100O are calculated, the point cor_ respondinr with these values can be found in either Fig_
ure 1 or
. If
2.
this point is at the right of the S-line (which gives the chosen value of the sdbility ratio), the soil p."r"rrr." q:"TI.. The next greater footing given by the O_tine (in Figure 1, the dotted D-line foi Jqra.e ftotings turn_ ing on A-A-axis only, and the solid D-line for" sqrare footings turning on A-A-axis as well as on B-B_axis) just above the point will take the loads without exceedingihe
allowable soil pressure. . If-the point is at the left side of the S_line, the size of the footing can be determined as follows: . Square. footing turning on the A-A-a_xis only. The footing given by the dottid D-line just above the point will be correct for soil bearing conditions, but this foot_ ing will not fulfill the stability ionditions. Draw a straight line through the origin and through the point (mIn_ tioned above) to the intersection with the S_line which gives the chosen stability ratio. The next greater footing given by the dotted D-line just above tte intersectioi will fulfill stability as well as soil bearing conditions. _ Square footing turning on the A-A-a:xis or on the B_ B-axis.. The footing given by the solid D-line just above the point is correct for soil bearing conditions. To find
footing which fulfills stability conditions, draw a origin and through ihe point to the intersection with_the the S-line which gives the chosen value of the stability ratio. The footin[ given by the dotted D-line just above the intersectioi is correct for stability conditions. The greater footing found from both cases will be correct for stability u, *"ll as for soil bearing conditions. Octagonal footing. This problem can be handled in _ the
.
straight line through
the same manner ar u rqrr." footing turning on an A_A_ axis only (see above) using FigurJ2.
Footing Weight. Before starting calculations, the weight of the footing must be estimated. (To simplify this e;i_ mate, the weights per inch depth of squaie_and octa_ g_onal-shaped ,pedestals or pads are givln in Table 1) . The-weight of reinforced concrete is tiken as 150 lbs. pir cu.. ft. The size of spread footing can now be found Uy itre
following procedure: 1. Fix the size of the pedestal in ft. and choose the shape. 2. Fix-the depth of the foundation below grade and the level above grade. This gives the totat heiglit of the foun_
dation. 3. Estimate pad thickness
in
inches.
4.
.F,stimate pedestal height in inches (which is the total height of foundatio, *irrrr. the estimated thickness of pad). 5. Using Table 1, calculate the weight of the pedestal in
kips.
6. Estimate the shape and size of the pad in ft. 7. Using Table 1, calculate the weight of the pad in kips. 8. Estimate backfill weight in kips. 9. Add 5, 7 and B for an estimated total weight of the
foundation.
10. Deterrnine the size of pad with Figure I or with Figure 2, taking into account the weighl of equipment and the overturning moment about thJbase of ihe foot_ ing. Divide the vertical load and the overturning moment by the soil pressure coefficient N.
Exomple
l.
Size
a
spread footing
for a
tower
, 54 ft.
high, skirt 5 ft. in diameter, empty weight 37 kiis, operating- weight 107 kips, test weighi lg7 kips. The Iateral Ioad due to wind is 10.5 kips. Tlie frost [nL is 3 ft. below grade. The allowable soil pressure at + ft. below grade is 2000.1bs. p"I sq.ft. The weight of soil is 100 lbs."per cu. ft. The minimum stability ritio is 3.5 For tower under test condition, the wind load is not to be taken into ac_ count. The solution:
o The pedestal will be 6 ft. wide, since allowance must be made for anchor bolts, etc. The shape will be octa_
gonal.
o The base of the foundation will be 4 ft. below grade (1 ft. below frost line). The level above grade is 1 ft.
so that the total height
o The pad
thickness
inches.
of the foundation is 5 ft.
will be assumed to be 2 f.t.:24
o The height of the pedestal is 5 ches.
o The-weight of the (see table).
pedestal is 36
o The shape of the pad sumed size of 16 ft.
..T1"
. Thg w91ght of X 0.i : 36.4 kips.
2
: 3
ft.
:
36
X O.Z7l:13
in_
kips
will be octagonal with an as_
weight of the pad is 24
table).
-
X Z.AS:
backfill is 0.8284
63.6 kips
(see
X (16,-6,) X Z
9l
GRAPHS SPEED SPREAD FOOTING DESIGN
o Total
weight of the foundation is 13
+
..
63'6
*
36'4
:
113 kips.
o The overturning moment about base of footing is 10'5 X ( (54 --2) + 5) : 340 ft.-kiPs'
: Empty weight of tower plus foundation is 37 * 113 150 kips. Operating weight of tower plus foundation is 107 + 113 : 300 kips' Test weight of tower plus founda-
tion is 187 + 113 : cient is 2000 M1000
:
-:-
340
v1000
:
:-
2
:
M1000
v1000
:
:
:
2. 170
150 -:- 2 :
M1000
v1000
250 kips. The soil pressure coeffi-
1000
220
ft' The minimum stability footing can turn on an A-A-axis only'
weight of soil is 100 lbs. Per cu.
.
ratil is 1.5. The
The solution: o The pier will be 1 ft. wide and square shaped' o The base of the foundation will be 4 ft. below grade (1 ft. below the frost line). o The total height of the foundation will be 5 ft'
o The pad thickness is assumed to be B inches' o The height of the pier is 5 ft. minus B inches :
52
inches.
ft.-kips; 75 kips
(empty).
170 ft.-kips; 1 10 kips (operating)
-- 2 :
.
0; 150 kips (test).
o Weight of the pier is 52 X 0.050 -- 4 -: 0.65 kips. o The pad wilt be square shaped, and is assumed to be 4.5 ft. wide. o Weight of the pad is B X 0.254 :2.032 kips. o Weight of backfitl is (4.5'-1') X (+ tt.-8") X
:
5.075 kips.
Draw a line in Figure 2 through M1000:170 ft'-kips and a line through v1000: 110 kips and find point 01' This point is at the right side of the S : 35 line, so the stabiliiy is all right. Point Or is between D : 15' - 6rr 16 ft' line and the D I 16 ft. line. The pad should be v1000.: for operating conditions. Drarv a line through 75 kips and-find point Or. This point is at the left side of the S : 3.5 lini, so when the tower is empty, the sta' bility condition governs. Draw the straight line "a" through the origin and point Oz and find point 03, which is the- intersection of line "a" and the S : 3'5 line' Choose the next greater pad size D : 16'. For test conditions, the pad ihould be 14 ft. wide. A 16 ft' wide octagon will be correct for soil bearing as well as for stability conditions.
0.1
Exompte 2. Size the footing for a pipe rack' The minimum vertical load is 8 kips; the maximum vertical load is 24 kips; the overturning moment at the top of the foundation is 9 ft.-kips; the iateral load at the top of the foundation is 3 kips' The anchor bolts are 5 inches, center to center. Top of the foundation is 1 ft' above grade' The frost line is 3 ft. below grade. The allowable soil pressure at 4 ft. below grade is 4000 lbs. per sq' ft' The
of the S : 1.5 line. The pad size should be 4'- 6" (use dotted D-lines) . For 6 ft.-kips and 4 kips, point 02 can : 1'5 be found. This point is at the ieft side of the S : D: 1' and line and betwe& the dotted D-lines, D with intersection the to +' 6". Draw a straight line "a" should pad the : For stability, 03)' (point 1.5 line the S be 4.5 ft. wide. A 4.5 ft. wide pad will be correct.
F. B. van Hamme is chief structural engineer for Fluor-Schuytvlot N.V., Haarlem, Holland (a subsidiary of the Fluor CorPoration,
N:
32 kips'
As in example 1, point 01 in Figure 1 can be found for 6 ft.-kips and 8 Lips. This point is at the right side
Exomple 3. The footing for a pipe support is square ,hrped and 5 ft. wide. the totul load * foundation is
solution:
Ltd.). He holds a civil engincer's desree front the Technical Univeisitv at Dclfr. Holland. Following giaduation, he enlisted in the Dut& army for two years. At the end of the tour of dutY in 1957, he left the army as a lieutenant of the
92
* foundation is 24 -l B : : 4000-+- 1000 4. M1000 : 24 -- 4 : 6 ft.-kips. v1000 : 16 -- 4 : 4 kiPs (minimum)' v1000 : 32 -- 4 : B kiPs (maximum)'
kips. Maximum load
16 kips minimum and' 32 kips maximum' The overturning moment about the base of the footing is 24 ft'-kips' The M1000 : 6 ft'-kips; v10O0 soil pressure coefficient N : 4. : 4 kipt (minimum) ; v1000 : B kips (maximum)' This footing can turn on an A-A-axis, as well as on a B-B-axis' Minirium stability ratio is 1'5. Check this footing' The
About the Author
Royal Engineers. In 1958, he joined Schuytvlot as a technical assistant, and was later promoted to Principal design engineer and then to his present posrtron.
o Total weight of the foundation is 0'65 + 2'032 + : 8 kips. o The overturning moment about the base of the footing is 9.0 + 3 X 5 : 24 ft.-kips. . Minimum vertical toad * foundation is B * 8 : 16
5.075
van Hamme
Find point O, in Figure 1 for 6 ft.-kips and 4 kips' Ifhis point is between the solid D-lines D : 4'- 6" and D '5 ft., ,o, for soil bearing conditions, the footing must be : 1'5 line' 5 ft. wide. Point O, is at the left side of the S the footstability, For Os. Draw Iine "a" and find point point Find DJines)' (use dotted wide ing can be 4'-6" for and bearing soil For B kips. and 6 ft.-kip's fo. OI stanitty conditfons, the footing can be 4t - 6" wide in this case (use solid D-line) . Conclusion: the designed 5 ## ft. wide square footing is correct.
Use Graph to Analyze Pile Supports Circular reinforced concrete beams, piles or column foundation supports in poor soil can be easily analyzed for flexure and axial tension
tended into the slab to fix the pile at the base of the slab. Soil conditions were found by analysis to provide little lateral support to the pile and therefore it will be neglected, thus providing an additional factor of safety. Concrete fr' : 5,000 psi, z : 7, f" : 20,000 psi, ,4, : six No. B bars equally spaced on 5.0 inch radius. The flexural moment in inch pounds : 60 (4,000) : 240,000 in.-lbs. a in inches
Andrew A. Brown, Olefins Div., Union Carbide Corp., South Charleston, W. Va.
R:5inches.
:ffi:
elr:!-:
1.42.
/n.\
in.
10
pn
r:
7 inches.
:ryH:
.276.
TuB or,srcw of rectangular and circular reinforced concrete beams that are subjected to both bending and com-
SM:24 ltn\-,,)n:24 3(.216) r:2.03.
pressive forces has been thoroughly discussed in manuals and textbooks. But very little can be found that is helpful
A tabular form is set up as Table 1 and various fr values are used until e f r approximate s 1.42 by Equation B. Then
in working out permissible capacities for circular sections r'vhich are subjected to an axial tensile force. The purpose here is to create more interest in this phase of design. fhis is not considered the final answer to these problems, but a step toward a more expeditious and rational approach.
Curve CM
It
.25 .26
4.8
.255
5.0
CM+
SM
SM
8.3 8.3 8.3
13.1
COS
a
t-2K
Then P
:
vs
- v, and. Y,"*Y" : Y s-
f c
.50 .48 .49
10.
u.
To determine the force in the concrete let f", equal the 1.42
CV
From
ln
sy*
Curve
SY.CV
t6 (sv - cv
2.03 2.03 2.03
1.015
.38 .42 .40
.635
LO.2
.555
8.9 9.5
3*
8.3. z pn ,r
method to analyze it. The derivation of equations is based on the assumption that the stresses vary directly as the distance from the neutral axis of the pile or as a straight line as shown by Section X-X, Figure 2. The symbols used are those found in concrete design manuals or textbooks. It is noted that by taking moments of the forces acting on the concrete and reinforcing bars about the diameter Y-Y that M : Pe : M" * M,.By summation of the forces acting on Section X-X we haveV, - V" - P : 0.
q k vqlue qnd mqke iriol cqlculqtions until e/r opprooches
From
Assume
(+) ":
Derivolion of Equotions. Figure 1 shows a typical foundation and the outermost pile on the windward side of the foundation. The pile is assumed to be subjected to a tensile force and thr-r"rst. Therefore, a need exists for a
Exomple: As an example, the following design data of a reinforced concrete pile will be used: Outside diameter : 14* inches. Thrust applied by wind is 4,000 pounds, The maximum uplift is 24,00:0 pounds. The pile is considered fixed 10 feet below the bottom of concrete mat, and the reinforcement is ex-
-[55urng
.216
the unit stresses are computed by Equations 9 and
Pile-Supported Foundqtions. The many factors which affect plant locations often result in the selection of plant sites on mud flats or deltas. Soil bearing values u.e low there and foundations must be supported on piles. In many instances, ta1I structures which are in the hurricane or seismic belts require pile foundations that are capable of resisting thrust and uplift. Economy and other considerations dictate this requirement.
TABTE f
x
.975 .995
.595
elr
: CM+SM 16 --(sc-cv) I.28<7.42
7.52>7.42 1.40
0K
SY* - (3ft pn) (1.-2h) 96hM
(CM
+
SM)r3
96 (.255) 240.OO0 13.3 (7)3
:
1,290 psi
nf.
LR
* r (7-2k)l _ 2hr
7 (1,290) t5
+
7 (1-.4e)I
2 (.255) 7
:
21,700 psi
93
USE GRAPH TO ANALYZE PILE SUPPORTS
..
The moment of the force in the concrete about the Y-Y axis on the same elemental atea: rJM": xdV'' Since x : r cos { and using dV " above the total moment
.
21"'" M-:, . [i...r-cosa) (1-cosa) ' Jo
F
rinzScosQdQ
I,L
ut co J
= UJ IA (!lr
=xIJ
This becomes
*":+(
=
J
LL
a+cosasina-2cos3a
AT TOP OF PILES
E F
The force acting on the reinforcing bars is solved by .orr*rtirlg the rei"nforcement into an annular ring of the center area of concrete at rR distance l;om "qrlrrut".ti of pile and of thickness f which equals T;E
z. o C) IL o SOIL PRESSURE NEGLECTED TO SOIL
If f"
CHARACTERISTICS.
MOMENT ON PILE =60(4000)=240p00 lN.- LB.
equals the intensity
distance oi R
=
-+
+
+
+ =
from the
"o, moment of the force on this area about the dM : f"R cos + dA* By simiiar triangles l" f" -:-
THRUST ON PILES=H/NUMBER 0F PILES=4,00018S. p
{
of stress on the area dA" a Y-Y ais, dV, : fttdA"'The same axis
Rcosd-/cosa r (1 - cosa)
?.4,oooLBs.(AxtAL TENstoN)
= TOTAL WEIGHT S = SECTION MODULUS OF PILE GROUP
rcoso
ITI
PIRALED CAGE
SECTION (USED
-
PILE
IN EXAM PLE
)
Fig. l-Typical foundation supported on piles'
on the elemental area dA"' Then dV" : and 2ydx:2r sin 0 (r0 d0) "' "' g triangles it c'an be seen similar By d'0. ot ii'-- , iir. (cosd-cosa) that/..':t"1_.oro-,
stress intensity dArfr', dA"f
:2ydxf
Substituting these values for dA"f "', rve have
v^-,,2t"' ["6oro 'c (l_cosa)Jo'
cosa) (sin2i,dQ)
By integrating and substituting 2ft for (1 total force in the concrete becomes
v"::;\--
f 12 / sinsa
94
sln 4 cos 2A
+'
-
_
-
cos
tr)
the
SECTION X-X AXIS
acos
)
(1)
Fig. 2-Plan and section of pile through
X'X
axrs'
= |?(o {coso sin oZ cii3o-srii CV =: 2 sinso*Jsino sincd+3sinn .ne2r coi2o'-3; -1 ^ C;; ^^
.rc -1gffi-141r3 _ 96kM
"
f, =nf, [n+r(t:Zt)] SM=24pr* SV
t$t
=3pn rr, coso
=3pn
-f
n(l-2k)
.35 .30 .25
40 50
70 .90
t.0
r.5 2.0
CV
3.0 4.0
5.0 70 9.0 tO.O t5
20
CM
Fig.
3-A
plot of CM and CV
Then by substitution 17 7*
2f.tR _ ----=----; r(I-c(
/
Jo
and
. After integrating between the limits and 1- for Rt and Zk for I - cos a.
(R.or e r cosa) dE -
f. ?nrzr cosa
2f"tRz ru ,,,.--.......-. " - / (l
th-e
curves.
f180" l,^
-cosa) ./0,^totp-rcosa)
substituting
bn12
frsu"
,
30 40 s0
'":----TE cos@dd
moment of the forces in the reinforcement, about X_X
axls-
r
Rz r and,M^_f"pn -
4k-
r
(3)
(4)
By subtractinq. Equation 1 from 3 and multiplying by the followine is obtained: r (V"-V") -
A.NDREw^
Ta -7- I o:: =:: _ -
(sin3 a
\-,-
+
: f^r3 --l3 pnr cos 4 iT -
(2sirra
a*
L----r-
About the quthor A. Bnowrv, Ca,ptain, Ciuit
En_
gne-er Corps, U.S. Nauat Reserae, is a
enior.E ng ine er, O Ie fins Dt tilor,iliioi uarotd,e Uorp,, So. Charleston, W. Va,
S
ihi"i
M r. rousn' s pr o f e s si,onal ti_ -B clucl.cs seueral years in the-Brid,ge "*e "'De_
sin a cos2 a
2 3 sin q. cos2
-4COS
a_
)
3 acos
a)] (5)
This is equal to Pr.
pa-rtme.nt, State Road
-Comruision of West Virginia,, nnd he has seraed, as L seoeral citi,es. for During_his 72 years of actioe d;;;-;; the-U.S.,Naay iome of'his biilets w"eri: pubtic -q:ilq;.wo,tks officc,r. N"o"l it "s1;;;;;,' "riii,"'rt i;;*,pton Roads, v a., .N a u at A;r' s tatioi, *;;;; i" ii,Zn, and N au at Station, San Juan, pue.rto.ni"o; bii-gi\ili Construction
ortdge consultant
Qficer, Fifttt Naaat Distri,ct,
li"i"i;;";"; ind Operations i;;;;";; "fi';, pubtic Works iclr er'irit;;-;; Ait T r ain_ i?{ Pensacota, Fta. ie-si"""tir"i"ilni!t i, ;- ;;;;"; ii "t rnternationat !?"11, Association for Brii.re i"i n"r", SAME, ASCE, and has BScb a;;-"ct";;;;;" ;;,il'ilrest virsinia Off,i.cer, Eteaenth Naaat Di\tric;,-;;;" p tr and, Moint enan c_e_ S up
Uni,oet,sity.
TABTE
.10 .15 .20 .lo .30 .35 .40 .45 .50 .55 .60 .65 .70 .80
2-Cy
ond
C/Ul
volues for vorious k ond a volues
360 25'
45" 34'
530 08' 600
660 25' 780 25' 840 76'
90.
950 44' 7070 32'
7070
,
1130 35' 7200
.04 .11
.22 .38 .59 .85 1.18 1.56 2.00 2.50 3.06 3.68 4.36 5.10 5.87
.57 1.51
2.97
/.u5 9.67 12.58 15.77 18.85 22.03 25.77 28.03 30.65 32.93 34.79
95
The total moment of the internal
stresses Pe is
M" *
M
orEquation2*4:
M
:#[tz
t"{cososina-2
"
cossasinc)
-
cosasin3a* 24
32
pnn(#)]
(6)
Since
M"l M"
a F; :7:
1v"-v"1,
, .r !:Il rr(af
cosasina-2 cos3asina)
M
-
Equation 6
TEA;tt""5ro
:
;" -:
(a 1-cosasina "* -
2 coss
32 cosasin3 a
asina) -
@
e r
$
CM
cosasinsa*
32
5i1c
cos2 a
-
SM:24
{SM
(8)
sinc
-
2 cos3 a sin
a)
-
32 cos asin8 a
W
N, .s
3 a cos
c
in Equation 6:
t:G##*-Eq'e The unit
stress
f,:
in the reinforcement
(e)
'n
r R
M P
vc v8 Mc Ms
f"
becomes:
t,
-
Eq.
10
(10)
It
is important to observe that by a few changes in the formulas the graph car. be used to analyze columns or piles which are subjected to compressive forces and hori-
zontal thrusts. Equation B would be
96
aa p
nf.lR+r(t-2k)l 2kr
efr
: ##h
P + Vu - V" : 0orP: to P : V" - V" lor tension or
: 0:
*
Total vertical load on the pile group Number of piles in the Pile grouP Section moduhu of pile grouP
M' External moment N Number of bars
For values of k or a, CM and CV were computed (see Table 2) and the CM and CZ curves are plotted on Figure 3. Piles can be analyzed by use of Equation B. With the uplift, P, and the horizontal f.otce given, ef r can be computed. Values of k are then assumed and Equation B used until tJle e f r of the internal stresses is close to the e f r of the external load and moment. The unit stress in the concrete can be computed by substitution
F"
NOMENCLATURE
-F
SV:3 pn'T cosrr and CV : 2 sins a f 3 sin acos 2 a-
I
V" - V, as compared uplift on the pile.
R2
Pn'Ir
(7)
t6 (SV - CV)
where cos a
O"'(y))
/ nz\ *24 pn' (+ ) \r'/ 3 a cos a)]
obtained by CM :12 (a +
'r
insacos2a-3acosc)]
blc 12
-
which is
2
kr
2
a
2
q
t
at toP of Piles
Area of one bar in square inches Ratio of area of steel to area of concrete Ratio of modulus of elasticity of steel to that of the concrete Radius of concrete column or pile in inches Radius of dowel bar circle in inches External moment at the sections of pile fixity Total vertical load or axial tension on the pile or columas Total vertical force in the concrete or total stress Total vertical force or stress in the reinforcement Resisting moment of the concrete Resisting moment of the reinforcement Maximum unit stress in the concrete in pounds per square inch Maximum unit stress in the reinforcing steel in pounds
per square inch The distance to the neutral axis measured along a radius from the point of maximum stress in the concrete (&d ) The angle subtended by radii drawn from each end of the chord which forms the neutral axis The angle subtended by radii drawn from each end of Y axis any chord X distance from Y
a Eccentricity
ininclr.es
M/P
-
CM, CV, SM and SV are numerical coefficients ##
: II t
I
i I
T
i
I { q
J1
L
r'i
