brownell and youg baseplate design

8/3/2019 Brownell and Youg Baseplate Design

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Post-it~ Fax Note 7671 Date 1- II#of ..pagesTo C ~ f ''-, I/I! t= From . , l "r: F ", ,) ) r r ~~ (Co.fDept. Co.Phone # Phone #Fax # "-7 " o -. -0- P - ' ' ] ? Fax #.j :;. ; PTE R

D E S IG N O F S U P P O R T SF O R V E R T IC A L V E S S E L S

~ ertical vessels are normally supported by means of asuitable structure resting on a reinforced-concrete founda-lion. This support structure between the vessel and thefoundation may consist of a cylindrical steel shell termed a"skirt." An alternate design may involve the use of lugsor brackets attached to the vessel and resting on columns orheams. These more common designs for supporting verticalvessels will be described.10.1 SKIR T SUP POR TS FOR VE R TICAL VE SSE LS10.1 Skirt Thickness. Tall vertical vessels are usually

supported by skirts. Because cylindrical shells have all themetal area located at the maximum distance (for a givendiameter) from the neutral axis, the section modulus, Z, ismaximum, and the induced stress minimum for the metalinvolved. Thus the cvlindrical skirt is an economicaldesign for a support for a tall vertical vessel, The skirt isusually welded directly to the vessel. Because the skirt isnot required to withstand the pressure ill the vessel, theselection of material is not limited to the steels permitted bythe pressure-vessel codes, and structural steels with cor-responding allowable stresses may be used with someeconomv. The steels used in the design of flat-bottomedcylindrical storage tanks (see Chapter 3) are suitable for theskirts of vertical vessels. For structural loads a factor ofsafety of 3 based on the ultimate tensile strength is usuallyused, whereas a factor of safety of 4, is used with pressurevessels, Thus the allowable stress in the skirt is usually33:3So higher than that in the shell of a pressure vesselwhen the steels in each case have the same ultimate tensilestrength.The skirt may be welded directly to the bottom dished

head, flush with the shell, or to the outside of the shell. Ifthe skirt is welded flush with the shell, the weight of thevessel in the absence of wind and seismic loads places theweld in compression. On the other hand, if the skirt iswelded to the outside of the vessel, the weld joint is in shear;therefore this method is not so satisfactory, hut it is an easymethod of erection and is often used for ~mall vessels. 'There will be no stress from internal or external pressure

for the skirt, unlike for the shell of the vessel, but the stressesfrom dead weight and from the wind or seismic bendingmoments will be a maximum. The same procedure may heused for designing the skirt as for designing the shell, "ilichwas described in Chapter 9. :\ote: Subscript b refers tothe ba se of the skirt.

(9.17)

(9.75;

Dead-weight stress =a b ~n1f dt (9,fi

: :\ iax permissible compressive stress =callDw.1.5 X 106 _;-;-;- 1= V f'lt~ ~ -g - y.p. (9.81)r .

Max tensile stress =lmax = (j",b or I.b) ~ /db (9.78)Max compressive stress = Icmax = (fwb or !.b) + /db (9.80)

183


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1 8 4 Design of Supports for Vertical Vessel,

Wind

d --~--..Jfa)

L ,I~r.(d - kd) _ j _ _ _ _ _ ~ i< d -1(b)Fig. 10.1. S~etc" of loading of anchor bolls.

After the skirt and bearing plate have been designed, theskirt design should be checked for the reaction of the boltingchairs or ring. (See section 1O.lg.)10.1 b Skirt-bearing-plate and Anchor-bolt Design.

The bottom of the skirt of the vessel must be securelyanchored to the concrete foundation bv means of anchorbolts embedded in the concrete to prevent overturning fromthe bending moments induced by wind or seismic loads.The concrete foundation is poured with adequate rein-

forcing steel to carry tensile loads (143, 154, 155). Theanchor bolts may be formed from steel rounds threaded atone end and usually with a curved or hooked end embeddedin the concrete. The bolting material should be clean andfree of oil so that the cement in the concrete wil l bond to theembedded surface of the steel.When either a compressive or tensile load is applied to the

anchor bolts, the load is transferred from the steel throughthe bond to the concrete. Surface irregularities, bends, andhooks aid in transferring loads from steel to concrete. Asthe steel and concrete are bonded, the resulting strain is thesame for both the steel and concrete at the bond. Themodulus of elasticity of steel, Eg is about 30 X 106 psiwhile that of concrete, Ec, varies from about 2 X 106 to4 X 106 psi depending upon the mix employed. The ratioof these moduli is:

(l0.1)rewrit ing gives:But

andE - b _s -

Substituting gives:

But becuuse of the bond, fe = E"---j> f. induced =nIc induced (10.2)

Table 10.1 gives the value of fl as a function of the com-pressive strength' of the concrete, which in turn is a fundionof the mix used for the concrete.The bending moment and weight of the vertical vessel

result in a loading condition on the concrete foundationsomewhat similar to that in a reinforced-concrete beam.Figure 10.1 is a sketch representing the loading condition ofthe anchor bolts in the concrete foundation.Figure 10.1, detail a is a sketch showing the bearing plate

at the base of a skirt for a vertical vessel. In the calcula-tions it is assumed that the bolt circle is in the center of thebearing plate. Sometimes the bolt circle is made larzerthan the mean diameter of the bearing plate but should .betaken equal to it for simplicity of calculation since the erroris small and is on the safe side. The wind load and thedead-weight load of the vessel result in a tensile load on theupwind anchor bolts and a compressive load on the down-wind anchor bolts. If I e is the compressive stress in theconcrete, the induced compressive stress in the steel boltsin the concrete is given by Eq. 10.2. Thus n le is the inducedcompressive stress in the steel bolts on the downwind side,and f. is the maximum tensile stress on the upwind side.As the stress is directly proportional to the distance fromthe neutral axis, a straight line may be drawn from is ton le, as shown in detail b of Fig. 10.1. The neutral axis islocated a distance kd from the downwind side of the bearingplate and a distance (d - kd) from the upwind side.By similar triangles, we obtain:

----.h_ = nic(d - kd) kd

thereforenlc 1k =-~ =-.__ .~nle +f. 1+ U./nIc} (10.3)

Table 10.1. Average Values of Properties of ThreeConcrete Mi"es

Water Content I e ' n I eU.S. Gallons 28-day Ultimate 30 X 106 Allowableper 94-lb Sack Compressive Ec Compressiveof Cement Strength, psi Strength, psi7k ; 2000 15 8006% 2500 12 10006 3000 10 12005 37$0 8 1400


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where I.= maximum induced tensile stress in steel atbolt-circle center line on upwind side, poundsper square inch.I e = maximum induced compressive stress in eon-crete at bolt-circle center line on downwind side,pounds per square inchE.n =-Ec

If the maximum induced tensile stress in the volts, I.,andthe maximum induced compressive stress in the concrete,fe . at the center line of the bolt circle are known, k may bedetermined by use of Eq. 10.3.Taylor, Thompson, and Smulski (156) have expressed

the area of bolting steel in terms of an equivalent shell ofsteel of thickness tl having the same total cross-sectionalarea of steel as shown in Fig. 10.2.Referring to Fig. 10.2, we find that the location of the

neutral axis may be defined in terms of angle 0: (156).d/2 - kdcos 0: = =1 - 2kdj2

In the same figure consider an element of the halting steelmeasured by angle d f J . The area of this element is given by:

(10.4)

(10.5)The distance of this element from the neutral axis is

r(cos 0: + cos fJ )The maximum distance from the neutral axis for such

an element is:r(1 + cos 0: )

The stress in the element. is' . is directly proportional tothe distance from the neutral axis, and if the maximumstress is fs, f I = ~(cos 0: + cos 8)

.s r (I + cos a) (10.6)Multiplying the stress hy the elemental area gives the

elemental force in tension. dF r d F t = fsilr (cos o: '~ us 0) d O(1 + cos 0: ) (10.7)

The summation of the elemental furces on the halting steelin tension can be represented by tensile force F, located atthe center of tension and distance 1 1 from the neutral axis.Similarly the summation of the compressive forces on theconcrete in compression can be represented by a compressiveforce Fe located at distance 1 2 from the neutral axis.RELATIONSHIPS FOR THE TENSIOK SIDE. By integration

of Eq. 10.7 for the upper ami lower halves on both sides ofthe center line, we obtain:

P , ==! (cos 0: + cos O J d fJ" (I + cos 0: )

=.l)r [-~ ( 1 1 " - a) cos 0: + sin a ) ] (10.3)l+eoso:(10.9)

Skirt Supports for Vertical Vessels 185where C t is the term in the brackets and is a constant fora given value of k.To determine the distance h consider the element which

is located a distance of r{cos 0: + cos 9) from the neutralaxis. The moment of the force on this element times thislever ann is:

d M t = d F t r(cos 0: + cos fJ )f 1 [ (cos 0: + cos 9) ( (J + )J d O=e lr r cos cos 0:(1 + cos G I )= f.llr [reeos 8 + cos 0:)2] de(l+ cos 0: )

By integration,'1 - f i 2' ) 1 " (cos 0: + cos 8)2 '81. t - If ~ a" (l+ cos 0: )=21.llr2 [ ( 1 1 " - 0:) ('0520: + %(sin C i cos 0:) + ~ ( 1 1 ' - 0:)]l+cosa

(10.10)Dividing Ait by F, gives h.h=( " I I ' - C i) cos2 a + ~(sin 0: cos G I ) + ~ell' - a ) j r

( T o - a) cos 0: + sin e x(10.11)

(Kate that II is a constant for a given value of k. )RELATIONSHIPS FOR THE COMPRESSION SIDE. On the

compression side a similar procedure is used. A differentialelement of concrete and steel is considered having an areaof:

(10.12)where i2 =concrete width (exclusive of bolting steel, il )under the bearing plate, inches.The distance of this element from the neutral axis is:

r(cos fJ - cos 0:)

k-zd=tF cFig. 10.2. Plan view of 100ding on boiling steel and bea ring plote.


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1 86 Design of Supports for Vertical VesselsTable 10.2. Values of Constants C~,Ce z , andjas 0

Functionof k (156)k C o Ct z J~------~-,----~-.--

0 . 0 5 0 0 . 6 0 0 3 . 0 0 3 0 . 4 9 0 0 . 7 6 00 . 1 0 0 0 . 8 5 2 2 . 8 8 7 0 . 4 8 0 0 . 7 6 60 . 1 5 0 1 . 0 4 9 2 . 7 7 2 0 . 4 6 9 0 . 7 7 10 . 2 0 0 1 . 2 1 8 2 . 6 6 1 0 . 4 5 9 0 . 7 7 60 . 2 5 0 1 . 3 7 0 2 , 5 5 1 0 . 4 4 8 0 . 7 7 90 . 3 0 0 1. 5 1 0 2 . 4 - 4 2 0 . 4 3 8 0 . 7 8 10 . 3 5 0 1 . 6 4 0 2 . 3 3 3 0 . 4 2 7 0 . 7 3 3O . W O 1 . 7 6 5 2 . 2 2 ! 0 . 4 1 6 0 . 7 8 4o . l 5 0 1 . 8 8 4 2 . 1 1 3 0 . 4 0 4 0 . 7 8 5o . 5 0 0 2 . 0 0 0 2 . 0 0 0 0 . 3 9 3 0 . 7 8 60 : = ; 5 0 2 . 1 1 3 1 . 8 3 4 0 . 3 8 1 0.7850 . 6 0 0 2 . 2 2 4 1.765 0 . 3 6 9 o . 7 8 !

The maximum distance from the neutral axis for suchan element is:

r(l - cos a)The stress I e ' in the element is directly proportional to

the distance from the neutral axis, and if the maximumind uced stress is L ;i' -j r(cos B - cos a)c - c r(l - cos a) (10.13)

The corresponding compressive (cornp.) stress, f's(eornp.),in the steel on the compression side (see E q . 1 0 .2 ) is:

I (cos B - cos a)f - n],.(comp.) - c (1 )- cos a ( 1 0 . H )The corresponding compressive forces in the element are

obtained by multiplying the elemental stresses by the ele-mental areas.

(10.15)

I [cose-cosa]dF e(5teel) = nf c dA , = fct1r dB1- cos ct (10.16)The total compressive force on the element is equal to

the sum of the above two equations, or

[COS e - cos a ]dFc(cotal) = ({z + ntl)rJc dOI- cos a

By integration.

l aa cos 0 - cos aF; =1 2 + ntl)rJc2 de

o I-coso:F - ( , )1[2(Sin a - a cos a) J- iz T il r o 1 - cos a (10.17)

( 1 0 . 1 8 )where Cc =he term in the bracket and is a constant for agiven value of k.To determine the distance l2 the same procedure is used

as for the tension side. The moment of the force on the

element times the lever ann is:diU r = dF e r(cos (J - cos a)

(cos () - cos a)~=((2 + llll)r"!e ~- ..-_~ dB(1 - cos a)By integration.

[" ~(. ',+ 1 1t + l 'j 2" COS- 0' - ;: sin a cos ct) 2a2 TI j) cr _ 1 - cos a( 1 0 . 1 9 )

Dividine Me by Fe gives 1 2 .

[ :l( . ) 1 JI_a cos- Ci - z SJIl a cos a "7" 20: (2 -. r)Sill a - a cos a ( 1 0 . 2 0 )(i. \ote that l2 is a constant for a given value of k.)The total distance between the forces FI and Fe i~ equal

to Il+ 12 This distance divided by d gives the dimension-less ratio, j.

- a) cos2 ~ + - h 1 1 " - a) + i ! : ~in a cos a]( 1 1 " - a) cos a + sin a1 l t D : - % sin a cos a + a cos" a ]+ 2 -_.. (10.21)sin a - a cos a

Referring to Fig. 10.2, we find that the distance from theneutral axis to the center line of the vessel is (d/2) (cos a)and distance z d is equal to:

dzd = 1 2 + 2 cos Ci ( 1 0 . 2 2 )

;.

1 [. + ( t a - isin a cos ct + a cos2 a ) J=" cos a - sin 0: - a cos a( 1 0 . 2 3 )

The quantities C ~, C o, j, and z are given in Table 1 0 . 2as a function of k.BOLTING AREA AND BEARING-PLATE WIDTH. Taking a

summation of moments about Fe (see Fig. 10.2) we obtain;Mwind - lV d,,z d - FJ d = 0

therefore,lfw;nd - lVdwzd

jd( 1 0 . 2 4 )

Substituting for F t by Eq. 1 0 . 9 we obtain:( 1 0 . 2 5 )

And .4. = 2 1 1 " r h ; therefore(10.26)


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Referring to Fig. 10.2 and taking a summation of verticalforces, we obtain:

(10.27)Substituting for F, by Eq. 10.9 and F e by Eq. 10.18, we

obtain:fsitrC t + Wau' - (1 2 + nfJ)rfcCe =0

Solving for 12 , we obtain:h= Wdu' + (Cd. - CcfcT l)rh

CeferThe total width of the bearing plate will he 1 1 + 1 2 (Eq.

10.~5 plus Eq. 10.28). Therefore

(10.23)

Width of bearing plate, I. = 1 1 + t2 (10.29)Nomographs for the solution of anchor-bolt problems by

the method of Taylor, Thompson, and Smulski have beenpresented by Gartner (233). An alternate procedure hasbeen presented by Jorgensen (234).DETERMINATION OF BEARIXG-PLATE THICKNESS. The

thickness of the bearing plate is determined by the com-pression load on the downwind side of the vertical vessel.The minimum required width of the bearing plate was pre-viously determined by use of Eq. 10.29. The maximumcompressive stress between the bearing plate and the COIl-crete occurs at the ouler periphery of the hearing plate.The induced compressive stress at the holt-circle center linewas determined by successive approximation in calculatingthe required width of bearing plate (see Eq, 10.29). Equa-tion 10.30 gives the relationsnip hetween the maximuminduced compressive stress at the outer periphery and thecorresponding stress at the Loll circle.

, (2kd + / 3 )fc(max induced) = (fe(balt circle induced]! 2kd (10.30)Although the compressive stress varies from the maximumgiven in Eq. 10.30 to a lesser valueat the junction of theskirt and bearing plate, the value at the bolt circle may beused for simplicity of calculation in determining the requiredthickness of the bearing plate.BEARING PLATES WITHOUT GUSSETS. A bearing plate

without gussets may be assumed to be a uniformly loadedcantilever beam with fc(max induced) the uniform load. Themaximum bending moment for such a beam occurs at thejunction of the skirt and bearing plate for unit circurnfer-ential Iength (b = 1 in.) and is equal to:

( l ) i,H (,"oxl = fa max bl ~ =2 (for b =1) (10.31)where I =outer radius of bearing plate minus outer radiuso f skirt, inches

The maximum stress in an elemental strip of unit widthis given by:

(for b =)where t4 =bearing-plate thickness, inches

Skirt Supports for Vertical Vessels 187Letting f(max) =(allowable) and solving for t4 gives us:t:

t4 = I V3 fc maxlfcallo-.r.) (10.32a)The thickness of the bearing plate, 14 as calculated byEq, 10.32a is usually rounded off to the next larger standardthickness of plate.BE."RING PLATES WITH GUSSETS. If gussets are used to

stiffen the bearing plates, the loading condition on the sec-tion of the plate between two gussets may be consideredto act similarly to that of a rectangular uniformly loadedplate with two opposite edges simply supported by thegussets, the third edge joined to the shell, and the fourthand outer edge free. Timoshenko (lOi) has tabulated thedeflections and bending moments for this case as shown inTable 10.3.Note in Table 10.3 that for the case where lib = 0 (no

gussets or gusset spacing, b = 0 ) the bending momentreduces to Eq. 10.31, and the thickness of the flange isdetermined by Eq. 10.32. Also note that when lib is equalto or less than the maximum bending moment occurs atthe junction with the shell because of cantilever action.If lib is greater than t, the maximum bending momentoccurs at the middle of the free edge.To determine the hearing-plate thickness from the bend-

ing moments, Eq, 10.,'33 may be used.i, = i 6 1 1 , ' 1 ( m a x l\I fallOW. (10.32b)

DESIGC'l PROCEDURE FOR BOl,TIKG CALCULATW"S xxnSIZIj';G OF BEARll\"G PLATE. The location of the neutralaxis is determined hy the ratio of induced stresses. as indi-cated by Eq, 10.3. Thus the determination of minimumbolting and minimum width of hearing plate requires suc-cessive-approximation calculations. The value of k deter-mines the constants C t, C o , i. and z, which in turn determinethe values of F, and Fe and their locations.As a first approximation in the determination of k, f. may

be taken as the maximum allowable stress in the boltingsteel, but fc should not be taken as the maximum allowablecompressive stress in the concrete since the maximum com-

Table 1 0 .3. M aximum Ben d in g Momen ts in a Bearin gPlate with Gussets (107)

(Courtesy of :\IcGraw-Hill Book Co.)lib Al ( x = b12) M ( X = b12)Xy= 1 Vy= l

----.------ ---- ---~~--o 0 -0 .500 foF73 0 .0078f ,b~ -0.428f,F7 2 o . 0293fY -0. 319fJ~% 0.055S


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;~, ,-. ~".~-

'Zt;~I!~


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r\

concrete. The proposed bearing plate under the skirt hasa 9 ft, 8 in. inside diameter and an outside diameter of11 ft, 8 in. The bolt circle is 11 ft, 0 in. in diameter andcontains 24 steel bolts 2H in. in diameter (from Table 10.4,the area per bolt"" 3.72 sq in). Assume that under oper-ating conditions the dead "eight of the tower is 600,000 lb.A high wind velocity develops a wind moment of 8,000.000It-lb. A continuous compression ring is used. FromTable 10.1, n = E. /Eo = 10; i.(allow.) for the structural-steel skirt is 20,000 psi. Determine the maximum inducedstress in tension in the bolts and the maximum inducedstress in compression in the bolts. Also determine themaximum compression stress in the concrete at the outer-most edge ofthe bearing plate on the downwind side and thewidth and thickness of the bearing plate.For first trial assume j, = 20.000.From Table 10.1, idmax) = 1200.

_ (11 ft, 3 in.) - (9 ft. 8 in.) _ 12 .tlPTOPo"ed) - - -- - 2 - m.

By Eq. ] 0.3. estimating fc(bolt circle) = 1000,1

k(approx.) = 1 + 1 0 = ------ = 0.333" 1 + 20,000! l ! e (10) (l0(0)By rearranging Eq. ] 0.30 and solving for !c(bolt circle) we

obtain:. . = PO(f [ .. (2)(0.333)(11)(12) l

ic(bBy Eq. 10.30(2kd + f a )!e(max induced) =(!c(holt circle induced)) -2k_d

= 335 (2)(0.32)(11)(12) + 1 2 )(2) (0.32)( 11) (12)3 " ( i 7 . 5 + 12) 965 .8;) _~ = pSI,7.;)


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1 90 Design of Supports for Vertical Vessels

Skirt Full-fillet weld%' gusset plate00 either sideof anchor bolts

3"x3"x~"Lor larger

fig. 10.3. Rolledangle bearing plate.

Determina L io n 'I t' bearing-plate thickness by Eq, 10.32 isas follows:

/-.- .. - - . --;-I, = I V ( . 1 f o jail"".)I = 11 ft,8 in. - 10 ft. 0 in..~ 10 in .

r(-;-;-(9-6\1 . ).. J, .t4 = 10 ,\!--- = 3.81m.'I : ;O.OO{) (without gussets)As this t nickness is considered to be excessive, the bearing

plate will he stiffened with 2 1 gussets equally spaced andstraddling the bolts.The gusset spacing, b. is

17.3 in.

I 10- = -- 0.58b 17.3InterpolaLing het.ween Vb = :':l and iib = .% from Table10.3 gives:

Alma>:= Aly = -0.26/


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Skirt Supports for Vertical Vessels 191TobIe 10.5. Maximum Number of Centered Chairs in

Various-sized Vessel SkirtsSkirt diameter, ft ~o. of Chairs

3 44 85 86 127 168 169 2010 24

P l an V ie w B-Bwidth of the plate. With this considerution the requiredbearing-plate thickness inside the chair may be calculatedby Eq. 10.37.

BI 61v!max, "--=--\ J (t. - bhd)fnllow. (10.37)B

rl;(" f il let weld

~ 1 ~ 4 "r-~~iI. IN~where is = bearing-plate width, inches

t4 = hearing-pia te thickness, inchesbh d =bolt-hole diameter in bearing: plate, inches

j "uuw. = allowable stress, pounds per square inchThe bending moment. in the hearing plate outside the

stiffeners (between chairs) may be controlling and can lwdetermined by use of Table 10.3. The thickness can hedetermined by use of Eq, 10.32b.EMPIRICAL DnrEXSIO'ilS FOR EXTERI'IAL CHAIRS. If the

number of bolts required exceeds the number given illTable 10.5, external bolting chairs may he used, as shownin Fig. 10.6. The proportions for the chair may be deter-mined empirically by the relationships p-iYcn in detail b ofFig. 10.6. :'{ote that the hole in the hearing plate is made

Section .4.-,4

--A

Samea s s ki rt

LWasher (thickness equalto bait dia meter)LA

Elevation

f ig. 10.5. Centered onchor- boll choir.at or near the bolt where the cross-sectional area is mini-mum. The moment is given by Eq. 10.36. r-ol t s ize + " ," 'minMin = Yo t s

Bolt' + 8"IbIh size -b r4 Gusse t plates - ifma x = - (10.36) Qo ~ " "1 T Bolt size + i'Y l*-here Alma" = maximum bending moment, inch-pounds

b = spacing inside chairs. inches (usually 8 in.) (a) ( MThe hole in the bearing plate reduces the effective beam Fig. 10.6. External boiling chair.


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19 2 Design of Supports for Verticol Vessels

Fig. 10.7. Vesse l .kirt with external boiling choirs.

larger than the hole in the top plate for ease in erection ofthe vessel.CALCULATION OF COMPRESSION-PLATE THICKNESS. The

maximum load on the compression plate at the top of anexternal chair occurs on the upwind side of the verticalvessel where the reaction of the bolts produces a compressionload. The compression plate may be considered to act asa rectangular plate bounded by the two gusset plates, theskirt, and the outside of the plates. The bolt load may beconsidered to be a uniformly distributed load acting over acircular area equal to the bolt area. The fact that thecompression plate is welded to the skirt and gusset platesas indicated in Fig. 10.7 provides additional rigidity onthese sides, which tends to compensate for the lack of sup-port on the fourth side. As an approximation the platewill be considered to act as a plate freely supported on foursides. Timoshenko (107) has developed the relationshipsfor a rectangular plate freely supported on four sides with aconcentrated load acting as a uniformly distributed loadover a circular area of radius e. In reference to the "Plan"view of Fig. 10.6 with y in the radial direction and x in thecircumferential direction, the maximum bending momentsM y and M x are given by Eqs. 10.38 and 10.39, respectively,

[ (2l sin " I r a ) 1

My =~ (l + p) In 'Ire I + I - l ' Y : ](10.3&)

[ (2 1 sin 1 r a ) " 'j

=~ (l+ ! J . ) In tre I + 1-[ (1 - ! J . - 1'2 ) : J (10.39)

where M u = maximum bending moment along radial axis,inch-poundslvl", =maximum bending moment along circum-ferential axis, inch-pounds

P =maximum bolt load on upwind side (see Eq,10.35) pounds

!J . =Poisson's ratio (0.30 for steel)In =natural logarithma = radial distance from outside of skirt. to bolt.circle, inches1 = radial distance from outside of skirt to outer

edge of compression plate, inchesb = gusset spacing, inchese =adius, of action of concentrated load, inches=one-half distance across flats of bolting nut,

inches1'1. 1'2 = constants from Table 10.6A comparison of Eqs. 10.38 and 10.39 using the constantsin Table 10.6 indicates that for (bll) = unity, M,. = M y,

and that for all cases in which (bll) is greater than unity,.M u is greater than M e and " . . J ~ is therefore controlling.After the determination of the size of the bolt and thewid th of the bearing pIa te and after the selection of theholt-circle diameter and gusset spacing, the dimensionsa.b,e, and I are fixed. The constants 1'1 and 1'2 may beevaluated by use of Table 10.6, and the maximum bendingmoments in the radial and circumferential directions maybe computed by use of Eqs. 10.38 and 10.39.For the case in which a is selected to be l/2 and M y iscontrolling, Eq, 10.38 reduces to:

p l 2 1 Jy =- (1+ ! J . ) In - + (1 - 1'1)4? r 'Ire (10,.10)To determine the maximum stress in the compression

ring a strip of unit width is considered. For this case,6iHy

fmax = t52

TobIe 10.6. Constonts for Moment Colculotion inCompression Ring (l07)

(Courtesy of :.\IcGraw-HilI Book Co.)bll 1.0 1.2 lA 1.6 1.8 2.0 o o . .1'10.5650.3500.211 0.125 0.073 0.042 01'20.1350.1150.0850.0570.0370.023 0Nole: for a billess than 1.0 invert bll and rotate axes 90.


11/15

-',_,-,r1

Or, iffx is assumed to ~fallow.r-+: -= ~

is = Y(6My/faUo\\'JI ' " , (10.41)where is = thickness of compression plate, inchesfallow.= allowable working stress, pounds per square inchlO.le Example Calculation 10.2, External-ckair Design.

An external chair will be designed for a column 8 ft, 0 in.in diameter having 12 bolts Ui in. in diameter with a cal-culated induced stress of 17,500 psi. The bolt-circle diam-eter is 3 ft, 6 in., and the outside diameter of the hearingplate is 9 ft, 0 in. The gusset height, h, is 12 in.By Fig. 10.6,

(Use %-in. plate.)6 = (%)(1% m.) = 0.515 in.A = 9 in. + (1~ in.) = 101i in.b = 8 in. + (lH in.) = 9H in.

By Table 10.4,root area of bolt, .4 0 = 1.405 sq in.

The holt load by Eq. iO.35 is:P = f.Ab = (17,500)(1.405)

By Fig. 10.6,a = (8 ft, 6 in.) - (3.~_0 in.) = 3 in.

2

24,600 Ib

I = (9 . ft, 0 in.) (8 ft. 0 in.)._:___.:_._:__---,--__:_=6 in.2From Table 10.4,

nut dimension across flats 2.375e=- 1.188 in.2 2The compression-plate thickness is:

~= 9.5 = 1.58I 6Interpolating from Table 10.6 gives:

Substituting in Eq. IO.to gives:P [ "I ly = _ (1+ Il ) In = - + 1 - 1'14 ? r er

= 24,600 [1.3 In ( (2)(6) ) -;- 1 - 0.134l4 7 r -i.iss+8200 in-Ib

Substituting into Eq. 10..11 witbfallow. = 17,500 psi gives:i /6My /(6)(3200) 6s = '\.}-- - \1 - 1. 72'1 / fallow. 17,500

Therefore use l%-in. plate for compression plate.lO.lf Continuous-compression-ring Thickness. Figure10.8 shows a sketch of a continuous compression ring used

Skirt Supports for Vertical Vessels 193as the upper plate of the bolting ring. Such a continuousring is preferred when the spacing of external chairs becomesso small that the compression plates approach a continuousring. As in the case of the compression plate the maximumload on a continuous compression ring occurs on the upwindside of the vertical vessel where the reaction of the boltsproduces a compression load on the ring. This load pro-duces a bending stress in the compresson ring. As in thecase of external chairs the vertical gusset plates transfer thiscompression load to the bearing plate.In determining the thickness of the continuous compres-sion ring the assumption is made that each section of thering between gussets acts as a rectangular plate bounded bythe two gusset plates, the shell, and the outer ring. Thebolt load will be considered to be a uniformly distributedload acting over the area of the bolt.Therefore the method used in determining the thicknessof the compression plates for external bolting chairs isapplicable. This method involves the use of Eq. 10.38,

10.39, or 10.40 and of Table 10.6 plus Eq. 10.41.CALCULATIO:'II OF GUSSET-PLATE THIcKJ'\Ess FOR COM-PRESSIOJ'\RINGS. If the gussets are evenly spaced alter-

nately between bolts, the gusset plate may be considered toreact as a vertical column. Normally the gusset is weldedto the shell, but no credit is taken for the stiffening effectproduced by the shell. The moment of inertia of the gussetabout the axis having the least radius of gyration is givenin Appendix J, item 1as:

I = U6a = ar 2 = U6r~12

Com press ionring

Fig. 10.S. Skirt with continuous cempre ion ring ond .trap.


12/15

194 Design of Supports for Vertical Vessels

(10.42)

where a '" area of cross section, square inchesr '" radius of gyration, inchest6 =gusset-plate thickness, inchesI =width of gusset, inches

Equation 4.21 may be used to express the relationshipor steel columns in which the value of (hlr) is from 60 to200.

P I B , O O Of = ;;-'" 1+ WIIB,OOOr2) (4.21)where h = height of gusset, inchesSubstituting Eq. 10.42 into this relationship gives.

13,000fallow. = 1+ (h? 'I-OO t ., (10.43)-;.J 6")

The allowable stress, fallow., in Eq, 10.43 must be:f - Bolt load _ Pallow. - l i 6 - ; ; - (10.44),

Substituting in EL[. 10.43 gives:Bolt load

li sI S , O O OI+ W/1500ts2)

IS,000li6~ - (bolt loud)t62 - h2(boJt load) =0 (10.45)1500Examination of Eq. 10.45 indicates that when the gusseteight, h, is small, the third term in the equation may beisregarded. In this case Eq. 10.45 reduces to the rela-ionship for straight compression without column action or

bolt load: " 6 = - 1 - S - , O - O - O [ - (10.46)Equations 10.45 and 10,46 are based on the asumption

hat the compression plate is sufficiently thick for the boltoad to be transferred to the gusset plates without intro-uction of eccentric action. The stiffening resulting fromhe welding of the compression pia tes and gussets to thehell introduces a margin of safety which justifies the aboveIf the gussets are not evenly spaced, an eccentric loadingill result in an induced bending moment. The thicknessf such gussets may be proportioned empirically, as in the-ase of gussets for external chairs.

(10.1-7)10.1 g Reoction of External Bolting Choirs and Compres-on Rings. The use of external bolting chairs or a compres-ion ring results in a loading condition that produces a reac-ion in the skin. This reaction, R, is similar to the shear

force, Qo (see (Fig. 6.3) produced in shells with closures.and the calculation may be treated accordingly.In Chapter 6 the following relationships were derived:

(6.75)

(dY) =-; _ ( 2BJ l o + Qo)dx :


13/15

Substituting- Eq. 10.53 into 10.55 gives:Par3w = _-mh (to.56)

Skirt Supports for Vertical Vessels 195This thickness can be reduced by increasing the gusset

height. Assuming a gusset height of 13 in. rather than 12in. will reduce t.he skirt thickness to:

t = 1.214(U)H = 0.926 in. or 1.0 in.10.1 i Thermal Stresses in the Skirt. For the case in

which the vessel is operated at a temperature considerablydifferent from atmospheric temperature, a thermal stressmay be induced in the skirt as a result of the temperaturegradient near the junction of the skirt and the vessel,TEMPERATURE GR.'lDlE:'IT IN SKIRT. To minimize the

temperature gradient in the skirt, the skirt may be insulatedboth inside and out. Skirts of vessels are insulated insideand out for fire protection when manways are cut into theskirt. The modulus of elasticity decreases rapidly withincreasing temperature above 600Q F with resulting loss inelastic stability. F. E. W olosewick (160) has given anapproximate equation for the skirts of vessels with 2 to4H in. of insulation both inside and out.To; =~Tv - 50) - 6.03ix - 0.: :!89x2

+ 0.009xJ - O.00007X4 (10.60)Differentiating with respect. .t.oz gives:dT x = _6.037 _ 0.578x + '0.027 J'~ - 0.00028x3dx (10.61)

Substituting Eq, 10.56 into Eq. 10.54 for w gives:f3 3i2Par2Alo = - _c _

6(1 - ~2)mh(10.57)

where To; = temperature of skirt at .1 ' distance below junc-tion of skirt and shell, degrees Fahrenheit

Tv =emperature of fluid in vessel bottom, degreesFahrenheit.

z =distance below junction of skirt and shell,inches

THERMAL EXPANSION. As an insulated vessel is broughtup to operating temperature, it wi l l undergo thermalexpansion. If there is no restraint to this expansion, nostress will be induced. The metal both in the skirt and inthe shell at the junction will have the same temperature.From the junction t.o the foundation a temperature gradientwill exist, which will tend to produce a varying thermalexpansion. At any given point in the skirt the radialthermal expansion, y, is proportional t.o the coefficient ofthermal expansion, 0;, the radius of the skirt, r, and thetemperature difference T", or

(10.62)

For a strip of unit width under flexure6M o f3 3Par 2

JuIlow. =r= - (1 - p.2)mhSubstituting for { 3 by Eq. 6.86 and solving for f gives:

(10.58)

For steel in which p. = 0.3,t=U6( Pa ) h rlim l 1 allow.

where i =kirt thickness required to resist reaction ofexternal chairs or compression ring, inches

r =adius of skirt, inchesm =2. 4 (see Fig. 10.6) or b s (bolt spacing)P = maximum bolt load, poundsa =adial distance from outside of skirt to holt circle,

(10.59)

inchesh =gus~et height, incheslO.lh Example Calculation 10.3, Reaction of 0 Bolting

Ring. The tower described in Example Calculation 10.1 isto be modified so that it has a bearing plate extending6 J ~ in. out from the skirt with a bolt circle 31~ in. outside theskirt. Twenty-four bolts 21 2 in. in diameter are to be usedwith a continuous compression ring, and the gusset heightis to be 12 in. Determine the required thickness of theskirt to resist the reaction of the bolting ring. The maxi-mum induced stress in the bolts is li,450 psi, and the maxi-mum allowable stress in the skir t is 20,000 psi.

a = 3_!, .Hn.m = rr(126.5) =16.6 in.

24fallOW. = 20,000 psi

h = 12 in.p = 17..150 X 3.i2 sq in. per boltr = 60 in.

By Eq. 10.59( P ) %= 1.76 __ a - rY.imhfnl loR.

Substituting gives:t = l.i6 [(li,450)(3.i2)(3.25)]% (60)li

(16.6)(12)(20,000)= 1.214 in. or l}~in.

where y = radial thermal expansion, inchesa =oefficient of thermal expansion inches per inch

per degree Fahrenheitr = radius of skirt, inchesT " =Tv - Tx) =emperature of vessel bottom

minus skirt temperature. degrees FahrenheitDifferenliating Eq. 10.62 with respect h z, the distance

along the skirt from its junction with ~ i.v vessel. gives:


14/15

196 Design of Supports for Vertical Vesselsbut dT t = d(T v - T",) = -dTx; therefore

dydx d (10.63)

STRESS FROM BE:"fDlNG MOMENTS AND SHEAR. Theterm dy /dx represent the slope of the skirt from the verticalas a result of thermal deformation. This deflection can becompared to the deflection dyt/dx for a cylindrical shelljoined to a flat-plate closure, shown in Fig. 6.3. At thejunction. where x = 0,

(6.75)

(dY) =-!_(2{3iH o + Qo )dx ",=0 2(3 DlSubstituting Eq. 10.62 into Eq. 6.75 and Eq. io.ss into

Eq. 6.76 gives:

(6.76)

ar er: ( d V ) I-- =~ =q (2(3Mo + Q o)dx dx x= o 2,B-DlClearing fractions on the right side of these equations gives:

arTl(2;33D J ) = ~{3"tl0 - Qoor dT I 2 ' --- (2# D1) = 2;31Ho -r Q o (10.64)dx

Adding the two equations gives:ar dT I arT 1(2;33D 1) + -- (2{J-D 1) =JM odx

therefore(10.65)

Substituting into Eq, 10.64 gives:

therefore(10.66)

At the junction of the skirt and bottom dished head T' =T v - T x . and T v =T r; therefore T ' = . And, therefore.

dT l (10.67)o = a d(3 Dl- dxand

dT lQo =-ex d{32 D l- (10.63)dxor

M = t; (10.69)ex dB Dl--, dx

andQ 2 sr,0= ex d# Dt-dx (10.70)

A comparison of Eqs. 10.69 and 10.70 indicates thatf J 1 V l o =Q o (10.71)

A.'{JAL THERMAL STRESS, jat, AT JU;XCTlO:X WITH SHELL.(See Eq, 6.122.)

fa ' =M oc = I M10 =-601 d{3Dl dT", I', I '2 ? (10.72), t i- dxCIRCU~[FERE(\"TIAL TIJER~[AL STRESS. J e t , Xl" JUNCTION

WITH SHELL. (See Eq, 6.125.)J e t = :{3d({3} V ~ o . _ -1 - Q o) +' ~6JI0 : t [2 . (l0.73)

But by Eq. 10.71 8;V10 =-Qo; thereforef - ~6Moct - 'T (10.74)

SHE.O\RTHERMAL STRESS, l,t, AT JV:";CTION WITH SHELL.(See Eq. 6.121.)1 ,t =iQ o =a: d .S 2 Dl dT": .

t 21 d x (l0.75)IO1i Example Calculation 10.4, Thermal Stresses.

Consider a vessel having a diameter of 130 in. and a skirtthickness of H in. insulated inside and out with the skirtsupporting a shell in which the bottom temperature is7000 F. Assume that the temperature distribution in theskirt is given by Eq, 10.60. Calculate the thermal stressesat the junction.

a =7.6 X 10-6 deg FE =25.5 X 106 psiJl =0.27

(from Fig. 3.6)

By Eq. 6.36. = 4 /3 (1 - , u _ 1 =.../3(1 --0.2(2 ) = 0 927j3 " r 2t2 \j (65)2(0.5)2 .-

By Eq. 6.15Dr - EI3 =25.5 X 106(0.5)3 =23.7 X 10~

- 12(1 - , u 2 ) 12 (1 - 0.272)The temperature gradient at the junction. x =0, by

Eq. 10.61 is:( d T x )dx x= o

Axial thermal stress:Substituting into Eq. 10.69 gives:

-6.037

, dT1 \ > 1 0 = -a : d# Dr_Xdx-(7.6 X 10-6)(130)(0.227)(23.7 X 104)(-6,037)

= 333 in -lb per in.


15/15

By Eq. 10.72f - I 6Afo I - ! 6(388) I iat -! t2 ,-! (0.5) 2

Circumferential thermal stress:By Eq. 10.70

9320 psi

2 dTxQ o = d /3 D-fix= (7.6 X 10-6)(130)(0.227)2(28.7 X 104)(-6.037)= -88.2 lh per in.

By Eq. 10.74f = f . l6Mo Ict t 2 [(0.27)(9320)J =2115 psi

Shear thermal stress:By Eq. 10.75

f3 Q. 3 -38.2 2 6 .of = 2" -( = ~--- = 64. pSI0.5

10.2 lUG SUPPORTS FOR VERTICAL VESSELSThe choice of the type of supports for a vertical pressure

vessel depends on the available floorspace, the convenienceof location of the vessel according to operating variables,the size of the vessel, the operating temperature and pres-sure, and the materials of construction.Brackets or lugs offerman:' advantages over other types

of supports. They are inexpensive, can absorb diametricalexpansions by sliding over greased or bronze plates, areeasily attached to the vessel by minimum amounts ofwelding, and are easily leveled or shimmed in the field..\8 a result of the eccentricity of this type of support.,compressive, tensile, and shear stresses are induced in tbe

wall of the vessel. The tensile and compressive forcescauseindeterminate flexural stresses which must be combinedwith pressure stresses circumferentially and longitudinally.The shear forcesact in a direction parallel to the longitudinalaxis of the vessel, and the shear stress induced by theseforcesis relatively 80 small that they are often disregarded.Lug supports are ideal for thick-walled vessels since the

thick wall has a considerablemoment ofinertia and is there-fore capable of absorbing the flexural stresses due to theeccentricity of the loads. In thin-walled vessels, however,this type of support is not convenient unless the properreinforcements are used or many lugs are welded to thevessel.If a vessel with lug supports is located out of doors thewind load, as well as the dead-weight load should be con-sidered in the calculation ofP. However, as lug-supported.vessels are usually of much smaller height than skirt-sup-ported vessels, the wind loadsmay be aminor consideration.The wind load tends to overturn the vessel, particularlywhen the vessel is empty. The weight of the vessel whenfilledwith liquid tends to stabilize it.The highest compressive stresses in the supports occur on

the leeward side when the vessel is full because dead loadand wind load are additive. The highest tensile stresses

Lug Supports for Vertical Vessels 197are set up on the windward side when the vessel is emptybecause in this case the dead load is subtracted from thewind load. Therefore the stresses on the leeward side arethe determining factor for design of the supports. Themaximum total compression load in P pounds in the mostremote column is (164):

P "" 4P",(H - L) + ~WnD bc nwhere P . " , =otal wind load on exposedsurface, poundsH = height of vessel above foundation, feetL=vessel clearance from foundation to vessel bot-

tom, feetDbc =diameter of anchor-bolt circle, feet

11 = number of supportsIW = weight of empty vessel plus weight of liquidand other dead load, pounds

(10.76)

10.2a Lugs withHorizontal Plates. Figure 10.9showsasketch of a vessel supported on four lugs, each lug havingtwo horizontal-plate stiffeners. Suchlugs are ofessentiallythe samedesignas that showninFig. 10.6 for external chairs,and the same design procedure may be used. This typeof lug uses to advantage the axial stiffness and strength ofthe cylindrical shell to absorb the bending stresses producedby the concentrated loads of the supports. Both the topand bottom plates should have continuous welds as themaximum compressive and tensile stress occurs in these twoplates, respectively. These welds and the intermittentwelds of the vertical gussets to the shell carry the verticalshear load. The load, P, on the column has a"lever arm, a,measured to the center lineofthe shellplate. This moment

IH

fig. 10.9.:d iffe n e rs a

Sketch of ve.. ,,1 on four-lug support, ....th hori~ontolplote

brownell and youg baseplate design

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