Skirt & Anchor Bolt BROWNELL & YOUNG Process Equip. Design [PDF]

Zitiervorschau

CHAPTER

DESIGN OF SUPPORTS FOR VERTICAL VESSELS

V

head, flush with the shell, or to the outside of the shell. If the skirt is welded flush with the shell, the weight of .the vessel in the absence of wind and seismic loads places the weld in compression. On the other hand, if the skirt is welded to the outside of the vessel, the weld joint-is in shear; thereiore this method is not so satisfactory, but it is an easy method of erection and is often used for small vessels. There will be no stress from internal or external pressure for the skirt, unlike for the shell of the vessel, but the stresses Plc wind or seismic bending from dead weight and from ,-_the moments will be a maximum. The same procedure may be used for~designing the skirt as for designing the shell, which was described in Chapter 9. Note: Subscript b refers to the base of the skirt.

ertical vessels are normally supported by means of a suitable structure resting on a reinforced-concrete foundation. This support structure between the vessel and the foundation may consist of a cylindrical steel shell termed a "skirt." An alternate design may involve the use of lugs or brackets attached to the vessel and resting on columns or beams. These more common designs for supporting vertical vessels will be described. 10.1 SKIRT SUPPORTS FOR VERTICAL VESSELS 10.1 Skirt Thickness. Tall vertical vessels are usually supported by skirts. Because cylindrical shells have all the metal area located at the maximum distance (for a given diameter) from the neutral axis, the section modulus, Z, is maximum, and the induced stress minimum for the metal involved. Thus the cylindrical skirt is an economical design for a support for a tall vertical vessel. The skirt is usually welded directly to the vessel.. Because the skirt is not required to withstand the pressure in the vessel, the selection of material is not limited to the steels permitted by the pressure-vessel codes, and structural steels with corresponding allowable stresses may be used with some economy. The steels used in the design of flat-bottomed cylindrical storage tanks (see Chapter 3) are suitable for the skirts of vertical vessels. For structural loads a factor of safety of 3 based on the ultimate tensile strength is usually used, whereas a factor of safety of 4 is used with pressure vessels. Thus the allowable stress in the skirt is usually 331 % higher than that in the shell of a pressure vessel when the steels in each case have the same ultimate tensile strength. The skirt may be welded directly to the bottom dished

Wind-load stress = Itob

15.89deffii 2

(9.20)

db2t Mu.

(9.17)

Irro2.1 Seismic-load stress = Li =

Dead-weight stress =

fdb

8CWH

(9.75)

"irr 2i

W at pr

(9.6)

Max permissible compressive stress = fcallow. 1.5 X 106 / v tyt, < 4 y.p. (9.84) r

Max tensile stress = f tmax = Max compressive stress = fcmax = 183

(fWb

Or Lb)

(fwb or fsb)

fdb

(9.78) (9.80)

184

Design of Supports for Vertical Vessels and E, = f es Substituting gives: nfc

fs

E, •

E,

But because of the bond, es = is induced = nfc induced

(b) Fig. 10.1. Sketch of loading of anchor bolts.

After the skirt and bearing plate have been designed, the skirt design should be checked for the reaction of the bolting chairs or ring. (See section 10.1g.) 10.1b Skirt-bearing-plate and Anchor-bolt Design. The bottom of the skirt of the vessel must be securely anchored to the concrete foundation by -Means of anchor bolts embedded in the concrete to prevent overturning from the bending moments induced by wind or seismic loads. The concrete foundation is poured with adequate reinforcing steel to carry tensile loads (143, 154, 155). The anchor bolts may be formed from steel rounds threaded at one end and usually with a curved or hooked end embedded in the concrete. The bolting material should be clean and free of oil so that the cement in the concrete will bond to the embedded surface of the steel. When either a compressive or tensile load is applied to the anchor bolts, the load is transferred from the steel through the bond to the concrete. Surface irregularities, bends, and hooks aid in transferring loads from steel to concrete. As the steel and concrete are bonded, the resulting strain is the same for both the steel and concrete at the bond. The modulus of 'elasticity of steel, Es, is about 30 X 106 psi while that of concrete, Es, varies from about 2 X 106 to 4 X 106 psi depending upon the mix employed. The ratio of these moduli is: Ec (10.1) n=

E

rewriting gives: . Ecn = Ec But Ec — ec

(10.2)

Table 10.1 gives the value of n as a function of the compressive strength of the concrete, which in turn is a function of the mix used for the concrete. The bending moment and weight of the vertical vessel result in a loading condition on the concrete foundation somewhat similar to that in a reinforced-concrete beam. Figure 10.1 is a sketch representing the loading condition of the 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 calculations it is assumed that the bolt circle is in the center of the bearing plate. Sometimes the bolt circle is made larger than the mean diameter of the bearing plate but should be taken equal to it for simplicity of calculation since the error is small and is on the safe side. The wind load and the dead-weight load of the vessel result in a tensile load on the upwind anchor bolts and a compressive load on the down-Wind anchor bolts. If L is the compressive stress in the concrete, the induced compressive stress in the steel bolts in the concrete is given by Eq. 10.2. Thus nf, is the induced compressive stress in the steel bolts on the downwind side, and fs is the maximum tensile stress on the upwind side. As the stress is directly proportional to the distance from the neutral axis, a straight line may be drawn from L to nfe, as shown in detail b of Fig. 10.1. The neutral axis is located a distance kd from the downwind side of the bearing plate and a distance (d — kd) from the upwind side. By similar triangles, we obtain: nh (d — kd) kd therefore 1 k = nf, nf, h 1+ (h/nh)

(10.3)

Table 10.1. Average Values of Properties of f hree Concrete Mixes Water Content Allowable U.S. Gallons 28-day Ultimate Compressive per 94-lb Sack Compressive Strength, psi 30 XEc106 Strength, psi of Cement 15 800 2000 12 1000 2500 10 1200 3000 6 8 1400 3750 5

Skirt Supports for Vertical Vessels where fs = maximum induced tensile stress in steel at bolt-circle center line on upwind side, pounds per square inch. fc = maximum induced compressive stress in concrete at bolt-circle center line on downwind side, pounds per square inch Ea IL — Ec If the maximum induced tensile stress in the bolts, L, and the maximum induced compressive stress in the concrete, fc, at the center line of the bolt circle are known, k may be determined by use of Eq. 10.3. Taylor, Thompson, and Smulski (156) have expressed the area of bolting steel in terms of an equivalent shell of steel of thickness- tt having the same total cross-sectional area 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 a (156). d/2 kd cos a = — 1 -- 2k d/2

d.48 = t1rd8

where Ct is the term in the brackets and is a constant for a given value of -k. To determine the distance /I consider the element which is located a distance of r(cos a A- cos 0) from the neutral axis. The moment of the force on this element times this lever arm is: dMt = dF1 r(cos a + cos 0) = Ail, [ (cos a + cos 0) r(cos 0 ± cos a ] dO (1 + cos a) , [r(cos 0 ± cos a)21 do = J stir (1 + cos «) j By integration, Mt ---- fettr 22

f ' (cos a A-- cos 0)2 (1 A- cos a) d° —

IL

cost a + *(sin a cos a) + 4-(a 1 + cos a

(10.4)

In tie same figure consider an element of the bolting steel measured'by angle.d0. The area of this element is given -by: (10.5)

185

) (10.10)

Dividing Mt by Ft gives 11. 11 —

[ (x• — a) cos? a -F- 4(sin a cos a) A- Or — — a) cos a + sin a

r (10.11)

The distance of this element from the neutral axis is r(cos a A- cos 0) The maximum distance from the neutral axis for such an element is: r(1 + cos a) The stress in the element, f8', is directly proportional to the distance from the neutral axis, and if the maximum stress is fe, r (cos a + cos 0) (10.6) f8' = fs r (1 + cos a)

(Note that 11 is a constant for a given value of k.) RELATIONSHIPS FOR THE COMPRESSION SIDE. On the compression side a similar procedure is used. A differential element of concrete and steel is considered having an area of: dA, = t2r dB

(10.12)

where 12 = concrete width (exclusive of bolting steel, 11) under the bearing plate, inches. The distance of this eleinent from the neutral axis is: r(cos 0 — cos a)

Multiplying the stress by the elemental area gives the elemental force in tension, dFt. dFt = Llir

(cos a cos 8) dB (1 cos a)

(10.7)

The summation of the elemental forces on the bolting steel in tension can be represented by tensile force Ft located at the center of tension and distance /1 from the neutral axis. Similarly the summation of the compressive forces on the concrete in compression can be represented by a compressive force Fe located at distance 12 from the neutral axis. RELATIONSHIPS FOR THE TENSION SIDE. By integration of Eq. 10.7 for the upper and lower halves on both sides of the center line, we obtain: Ft = fatir2 a

Concrete in compression

Steel in tension

(cos a + cos 0) dB (1 + cos a)

I 2 = fstir L + (Or — a) cos a + sin a)] (10.8) i cos a fstirCt

(10.9)

Fig. 10.2. Plan view of loading on bolting steel and bearing plate.

186

Design of Supports for Vertical Vessels

Table 10.2. Values of Constants Ct , C„ z, and j as a Function of k (156) C't Cc 0.600 0.852 1.049 1.218 1.370 1.510 1.640 1.765 1.884 2.000 2.113 2.224

0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600

3.008 2.887 2,772 2.661 2.551 2.442 2.333 2.224 2.113 2.000 1.884 1.765

0.490 0.480 0.469 0.459 0.448 0.438 0.427 0.416 0.404 0.393 0.381 0.369

0.760 0.766 0.771 0.776 0.779 0.781 0.783 0.784 0.785 0.786 0.785 0.784

element times the lever arm is: dM, = dF, r(cos 0 - cos a)

By integration, nil)fcr 22

Mc = (12

= (12 + n11)fcr22

The stress L' in the element is directly proportional to the distance from the neutral axis, and if the maximum induced stress is fc, (10.13)

The corresponding compressive (comp.) stress, f's(comp.), in the steel on the compression side (see Eq. 10.2) is: (cos 8 - cos a) (1 - cos a)

f a (comp.) = af,

(10.14)

The corresponding compressive forces in the element are obtained by multiplying the elemental stresses by the elemental areas. dFc(concrete)

f

dAc

f,t 2r

[cos 8 - cos al 1 - cos a

d8

( cos - cos a] de 1 - cos a

a cos a)

cos2 a -

4-a

1 - cos a

Dividing M, by F, gives 12 . -

L

a cos2 a - *(sin a cos a) + sin a - a cos a

(r)

(10.20)

(Note that /2 is a constant for a given value of k.) The total distance between the forces Ft and Fc is equal to 1 12. This distance divided by d gives the dimensionless ratio, j. j -

fc = f r(cos 8 - cos a) c r(1 - cos a)

(cos 0 - cos a)2 dB 1 - cos a 0

(10.19)

12

The maximum distance from the neutral axis for such an element is: r(1 - cos a)

(cos 0 - cos a)2 dO (1 - cos a)

= ±

11 + 12 d [ Or - a) cos2 a + 4(2r - a) + 4 sin a cos a ' Or - a) cos a + sin a ..1a - 2 sin a cos a ± a cos2 a1 (10.21) + 1sin a - a cos a I

Referring to Fig. 10.2, we find that-the distance from the neutral axis to the center line of the vessel is (d/2) (cos a) and distance zd is equal to: d zd = 12 - cos a 2

(10.15)

[

cos a +

(10.22)

(4a - 4 sin a cos a + a cos2 a)] sin a - a cos a

(10.16)

(10.23)

The total compressive force on the element is equal to the sum of the above two equations, or

The quantities Ct, Cc, j, and z are given in Table 10.2 as a function of k. BOLTING AREA AND BEARING-PLATE WIDTH. Taking a summation of moments about Fc (see Fig. 10.2) we obtain:

dFc(steel)

rife' dA s = nfclir

dFc(total) = (12 + nli)rfc

ros 8 - cos a] 1 - cos a

do

Mwind - Wdwzd - Ftjd = 0

By integration,

therefore

Fc = (t2

nt .j )rfe2

F, = (t

riti)rf,

[2(sin a - a cos a)]

= (12 + ril l)rf,Cc

m =

cos 0 - cos a dO 1 - cos a o 1 - cos a

Pt

(10.17) (10.18)

where C, = the term in the bracket and is a constant for a given value of k. To determine the distance 12 the same procedure is used as for the tension side. The moment of the force on the

Mwind

VVdwZd

jd

(10.24)

Substituting for Ft by Eq. 10.9 we obtairk: Wdwld 11 Mwind fsrCtjd

(10.25)

And A, = 2irrii; therefore A, = 27r [

Mwind

WdwZd

Ct fsjd

(10.26)

Skirt Supports for Vertical Vessels Referring to Fig. 10.2 and taking a summation of vertical forces, we obtain: Ft ± Wd w — F, = 0

(10.27)

Substituting for Ft by Eq. 10.9 and F, by Eq. 10.18, we obtain: fstirCt

Wdw

(12 + rdi)rf,C, = 0

Solving for 12, we obtain: 12

Wdw (Ctf, — Ce fen)ril Cc fcr

(10.28)

The total width of the bearing plate will be Li + 12 (Eq. 10.25 plus Eq. 10.28). Therefore Width of bearing plate, (3 = /1,

/2

(10.29)

Nomographs for the solution of anchor-bolt problems by the method of Taylor, Thompson, and Smulski have been presented by Gartner (233). An alternate procedure has been presented by Jorgensen (234). DETERMINATION OF BEARING-PLATE THICKNESS. The thickness of the bearing plate is determined by .the compression load on the downwind side of the vertical vessel. The minimum required width of the bearing plate was previously determined by use of Eq. 10.29. The maximum compressive stress between the bearing plate and the concrete occurs at the outer periphery of the bearing plate. The induced compressive stress at the bolt-circle center line was determined by successive approximation in calculating the required width of bearing plate (see Eq. 10.29). Equation 10.30 gives the relationship between the maximum induced compressive stress at the outer periphery and the corresponding stress at the bolt circle. fe(max induced) = (fc(bolt circle induced))

(2kd 13 2kd )

(10.30)

Although the compressive stress varies from the maximum given in Eq. 10.30 to a lesser value at the junction of the skirt and bearing plate, the value at the bolt circle may be used for simplicity of calculation in determining the required thickness of the bearing plate. BEART''G PLATES WITHOUT GUSSETS. A bearing plate witho ossets may be assumed to be a uniformly loaded cantilever beam with fc(max induced) the uniform load. The maximum bending moment for such a beam occurs at the junction of the skirt and bearing plate for unit circumferential length (b = 1 in.) and is equal to: / M(max) = fe max Oi 2 =

2

(for b = 1)

(10.31)

where I = outer radius of bearing plate minus outer radius of skirt, inches The maximum stress in an elemental strip of unit width is given by: f(max)

6M(max)

3fc max 12

bi 4 2

142

(for b =

where / 4 -- bearing-plate thickness, inches

187

Letting Amax) = f (allowable) and solving for 14 gives us: 14 = 1 V 3f c maxg(allow.)

(10.32a)

The thickness of the bearing plate, 14, as calculated by Eq. 10.32a is usually rounded off to the next larger standard thickness of plate. BEARING PLATES WITH GUSSETS. If gussets are used to stiffen the bearing plates, the loading condition on the section of the plate between two gussets may be considered to act similarly to that of a rectangular uniformly loaded plate with two opposite edges simply supported by the gussets, the third edge joined to the shell, and the fourth and outer edge free. Timoshenko (107) has tabulated the deflections and bending moments for this case as shown in Table 10.3. Note in Table 10.3 that for the case where l/b = 0 (no gussets or gusset spacing, b = co) the bending moment reduces to Eq. 10.31, and the thickness of the flange is determined by Eq. 10.32. Also note that when l/b is equal to or less than *, the maximum bending moment occurs at the junction with the shell because of cantilever action. If 1/6 is greater than 2, the maximum bending moment occurs at the middle of the free edge. To determine the bearing-plate thickness from the bending moments, Eq. 10.33 may be used. 14 -= -‘,/6.11(„„„)

(10.32b)

faiiou . DESIGN PROCEDURE FOR BoT:N c CALCULATIONS AND SIZING OF BEARING PLATE The location of the neutral axis is determined by the ratio of induced stresses, as indicated by Eq. 10.3. Thus the determination of minimum bolting and minimum width of hearing plate requires successive-approximation calculations. The value of k determines the constants C1, Cc, j, and z, which in turn determine the values of Ft and F, and their locations. As a first approximation in the determination of k, f, may be taken as the maximum allowable stress in the bolting steel, but fc should not be taken as the maximum allowable compressive stress in the concrete since the maximum corn-

Table 10.3. Maximum Bending Moments in a Bearing Plate with Gussets (107)

1/b

(Courtesy of McGraw-Hill Book Co.) x = b/2) ky fx =b/2) v

—0 .500fd2 0 — 0 . 4280 0 .0078f,b2 —0 .319f,t2 0 .0293f,b2 2/3 — 0.227012 0 .0558fcb2 . 119fc12 . 0972f,b2 0 .124012 0 .123M2 0 .131fcb2 2 --00112 125 5f:1/ 2 0 . 133M' 3 — 0 .125,02 0 . 133f,b2 b = gusset spacing (x direction) inches. 1 = -bearing-plate outside radius minus skirt outside radius (y direction) inches. 0

188

Design of Supports for Vertical Vessels Table 10.4. Bolt Data (157)

(Courtesy of Taylor Forge &Pipe Works) Minimum 8-thread Series Bolt Spacing* Bolt Standard Thread Radial Size No. of Root No. of Root Minimum Distance Pred Threads Area Threads Area B, ferred ;i" 13 0.126 No. 8 1y4" 3" 1 H6" 1%6 0.202 thread 1% 3 %" 11 3 1% %" 10 series 1% 2316 3 below 1" 1% 0.419 8 0.551 2% 3 1% 0.551 1" 8 1%" %," 1%" 1,%"

7 7 6 6

0.693 0.890 1.054 1.294

8 8 8 8

0.728 0.929 1.155 1.405

232 21,1s 3346 3%

1,%" 1%"

51.‘ 5 5

2"

432

1.515 1.744 2.049 2.300

8 8 8 8

1.680 1.980 2.304 2.652

3% 3% 4 4%

2%" 21. 4" 294" 3"

4% 43 3.020 8 3.423 4 4.2925% 3.715 8 5% 4 4.618 8 5 .259 4 6.324 8 5.621 614 center-to-center distance between bolts, inches

1%"

*

pressive stress in the concrete is developed at the outer periphery of the bearing plate rather than at the center line of the bolt circle. After evaluating k by Eq. 10.3, the minimum area of bolting steel required may be determined by Eq. 10.26 and the preceding relationships. This permits the selection of the number and size of bolts having sufficient root area to equal or slightly exceed the minimum required bolting area. Table 10.4 gives the necessary information for this selection. Usually the number of bolts selected is a multiple of four to permit greater ease in bolt layout. As(act.) = NAB

As(min)

(10.33)

where N = number of bolts A B = root area of bolt, square inches (Table 10.3) As(min) = minimum area of bolting steel, square inches (Eq. 10.26) As(act.) = actual area of bolting steel used, square inches The width of the bearing plate may be evaluated by use of Eqs. 10.25, 10.28, and 10.29. In the first trial the area of bolting steel is selected and the width of the bearing plate determined by use of the original evaluation of k. The values of the induced tensile

3 3

Edge Distance E

Nut, Dimension (across flats) 7 /g1/

Maximum Fillet Radius X

% 11-if 6 1%6 1%6

1346 1% 1%6 1%

ifs 24 % Xs

132 1% 1% 2

1% 1% 1% 134

1916 2 2916 2%

Hs

234 2% 2% 2%

1% 131 1% 2

2916 2% 21916 3%

91 % 91 1316

234 3346 3% , 3%

2%2% 2% 2%

334 3,1 434 4%

1316 1., 7,.6 %

9/ 916

946 %

6, 1/1 6

and compressive stresses in the steel and concrete based on this area and this width are next determined. An induced tensile stress in the steel based upon the first calculation of k may be determined by Eq. 10.34. fs(induced) = fs(first estimate)

As(min) A A 8 (act.)

(10.34)

An induced maximum compressive stress in the concrete also based upon the first calculation of k may be determined by use of Eq. 10.30. These values of fs and L may be used to obtain a more correct value of k by Eq. 10.3. If the new value of k differs appreciably from that originally calculated, a complete recalculation should be made using the constants Ct, Cc, j, and z based upon the new value of k. After consistent values of f„, fc, and k have been determined, the bolting design and bearing-plate width are established. The thickness of the bearing plate may then be established by use of Eq. 10.31. To complete the design of the skirt and bolting ring the compression ring or bolting "chairs" should also be considered. 10.1c Example Calculation 10.1, Bearing-plate Design.

A proposed fractionation column is 10 ft in diameter and 150 ft higliand rests on a foundation of 3000-psi-strength

Skirt Supports for Vertical Vessels concrete. The proposed bearing plate under the skirt has a 9 ft, 8 in. inside diameter and an outside diameter of 11 ft, 8 in. The bolt circle is 11 ft, 0 in. in diameter and contains 24 steel bolts 23i in. in diameter (from Table 10.4, the area per bolt = 3.72 sq in). Assume that under operating conditions the dead weight of the tower is 600,000 lb. A high wind velocity develops a wind moment of 8,000.000 ft-lb. A continuous compression ring is used. From Table 10.1, n = = 10; f“idit,„..) for the structural steel skirt is 20,000 psi. Determine the maximum induced stress in tension in the bolts and the maximum induced stress in compression in the bolts. Also determine the maximum compression stress in the concrete at the outermost edge of the bearing plate on the downwind side and the width and thickness of the bearing plate. For first trial assume L = 20,000. From Table 10.1, fe(n..) = 1200.

600,000 + 598,000 = F, = 1,198,000 lb The induced stress in the concrete at the bolt circle, L, based upon k = 0.333 may be evaluated by Eq. 10.18.

F, =-

k(approx,)

1 1 + .fa

nf,

1 20,000 1+ (10)(100.0) 0.333

By rearranging Eq. 10.30 and solving for fc(boit circle) we obtain: ic(bolt circle)

= 1200

= 1200

(2)(0.333)(11)(12) 1 12i (2)(0.333)(11)(12) 88 (88 + 12) = 1055 psi

To evaluate the induced stresses the ,.!onstantsare read from 'Pable 10.2. For k = 0.333 Ce

1,198,000 = [(11.785) ± (10 X 0.215)](5.5)(12)(1.588)h

f, = 818 psi Rechecking k by Eq. 10.3 gives:

Ft —

Al %mid — Wdwzd 8 X 106 — 6 X 105(0.431)(11) — (0.782)(11) jd

_8 X 106 — 2.85 X 106 .---- 598,000 lb 8.6

Ct — 2.405 z = 0.434

j = 0.782' 8 x 106 — 105(6)(0.434)(11)

Ft fs

(0.782)(11)

=

Ft = fstlrCt A (24)(3.72) 5d = 5(11)(12)

F, = 600,000 ± 596,000 = 1,196,000 lb 1,196,000 — 835 psi (11.785 + 2.15)(5.5)(12)(1.554) 1 17,400 1+ (10) (835)

598,000 = L(0.215)(5.5)(12)(2.376)

1.3 = 17,700 psi

1 —0.324 1 ± 2.08

k = 0.32, approirnately, by interpolation The rechecks of fs, fc, and k are in sufficient agreement for design purposes. To check maximum compressive stress in bolts and concrete, Eq. 10.2 is used.

Es — L = iofc = 8350 psi Ec

By Eq. 10.30 fc(max induced) = (fc(bolt circle induced))

0.215 in.

596,000

596,000 — 17,450 psi (0.215)(515)(12)(2.405)

fs(comp ) =

The induced stress in the steel L based upon k = 0.333 may be evaluated by Eq. 10.9.

1 — 0.317 1 ± 2.16

= 1.554

—

The tensile load may be calculated by Eq. 10.24.

1 17,700 1+ (10)(818)

Rechecking constants and stresses gives the following. From Table 10.2 for k = 0.317,

= 2.376

0.782

nli)rfeCc

12 = 13 — ti = 12 — 0.215 = 11.785 in.

1.588

z = 0.431

(12 +

But

(11 ft, 8 in.) — (9 ft, 8 in.) = 12 in. 2

By Eq. 10.3, estimating fc(bott circle) = 1000,

Wd. — F, = 0

Ft

k t(proposed) =

189

The compressive load may be calculated by a summation of vertical forces using Eq. 10.27.

(2kd + 13 9 kd

835 t(2)(0.32)(11)(12) + 12 \

(2)(0.32)(11)(12) 77.5 ± 12) — 965 psi = 835 ( 77.5

190

Design.of Supports for Vertical Vessels By Eq. 10.320 Full-fillet weld

Skirt

t4 = V6(11,700/20,000) = 1.875

)g" gusset plate on either side of anchor bolts

Therefore, use 1%-in. plate. 10.1d Practical Considerations in Designing Bearing Plates.

%.* fillet weld

rxrxV.Z. or larger Z9' Alla' A

Bolt size + Fig. 10.3. Rolled-angle bearing

pica*.

Determination of bearing-plate thickness by Eq. 10.32 is as follows: l 4 = l 03f c/fallow.) 11 ft, 8 in. — 10 ft, 0 in. — 10 in. 2 t 4 = 10 V(3)(965) — 3.81 in. 20,000

(withe.it gussets)

As this thickness is considered to be excessive, the bearing plate will be stiffened with 24 gussets equally spaced and straddling the bolts. The gusset spacing, b, is b=

/ b

7r(11)(12) = 17.3 in. 24 10 = 0.58 17.3

Interpolating between l/b = 3 and 1/b 10.3 gives:

from Table

Mm„„ = M, = —0.26A/2 = (-0.26)(965)(100)

•= —25,100 in-lb

ROLLED-ANGLE BEARING •PLATE. If the vertical vessel is not very high and a skirt is used to support the vessel rather than legs, lugs, or columns, a simple design may suffice for the bearing,plate. If the calculated thickness of the bearing plate is 3z in. or less, a steel angle rolled to fit the outside of the skirt may be lap welded as shown in Fig. 10.3. SINGLE-RING BEARING PLATE. If the required bearingplate thickness is in. to in., a design using a single-ring bearing plate may be employed, as shown in Fig. 10.4. If the bearing-plate width is less than 5 in. and the thickness less than 3 in., the rolled-angle design (Fig. 10.3) will probably be more economical. CENTERED CHAIRS. If the required bearing-plate thickness is 3/4 in. or greater for the design shown in Fig. 10.4, a bolting "chair" can be used to advantage. Figure 10.5 shows a typical design for a centered anchor-bolt chair. Although the number and size of bolts required should be checked for each design, Table 10.5 gives some typical values of the maximum number of chairs usually inserted in a vessel or skirt of a given diameter. In checking the bearing-plate thickness for a :,-entered chair the plate inside the stiffeners may be considered to act as a concentrated loaded beam with fixed ends. The concentrated load, P, is produced by the bolt and is equal to maximum bolting stress times the bolting area, or fs Ab .1) (10.35)

where h

= maximum induced stress in bolting steel

Ab = root area of anchor bolt, square inches (The values of A and Ab are those determined in an

earlier section.) P = maximum bolt load, pounds

The maximum bending moment in the bearing plate inside the chair occurs at upwind dead center and is located

By Eq. 10.32b t 4 = V6(25,100/20,000) =- 2.75 in. Further reduction, in bearing-plate thickness could be realized if the gusset spacing were decreased by using 48 gussets. For 48 gussets, gusset spacing, b is: b—

7(11)(12) = 8.65 in. 48

10 = = 1.255 8.65 b

Skirt

gusset plate on either side Of anchor bolts

54•• fillet Bolt size +

weld

/

Interpolating again from Table 10.3 gives: Minas = My — 0.12ifei2

—11,700 in-lb

/

A1111// A 2. 5" minimum

Fig. 10.4. Single ring beam, plate with gussets.

Skirt Supports for Vertical Vessels

WI

Table 10.5. Maximum Number of Centered Chairs in Various-sized Vessel Skirts Skirt diameter, ft No. of Chairs 3 4 4 8 5 8 6 12 7 16 16 8 20 9 24 10

width of the plate. With this consideration the required bearing-plate thickness inside the chair may be calculated by Eq. 10.37.

Plan View B-B

6 /1/„. — bhdXfatow.

14 =

(10.37)

where is = bearing-plate width, inches 14 = bearing-plate thickness, inches bhd = bolt-hole diameter in bearing plate, inches fano, = allowable stress, pounds per square inch The bending moment in the bearing plate outside the stiffeners (between chairs) may be controlling and can be determined by use of Table 10.3. The thickness can be determined by use of Eq. 10.32b. EMPIRICAL DIMENSIONS FOR EXTERNAL CHAIRS. If the number of bolts required exceeds the number given in Table 10.5, external bolting chairs may be used, as shown in Fig. 10.6. The proportions for the chair may be determined empirically by the relationships given in detail b of Fig. 10.6. Note that the hole in the bearing plate is made

Bolt size + Washer

1

/4-or greateF--1— T

/ Bolt size +

6 Section A-A

Washer (thickness equal to bolt diameter) Bolt size + 9' Bolt size + 1/4"

Elevation

t5

Fig. 10.5. Centered anchor-bolt chair.

near the bolt where the cross-sectional area is minimum. The moment is given by Eq. 10.36.

at or

Pb

Mmax = — 8

(10.36)

where Mmax = maximum bending moment, inch-pounds b = spacing inside chairs, inches (usually 8 in.) The hole in the bearing plate reduces the effective beam

Bolt size + V min Min = %to Bolt size + 8` Gusset plates to

Qo

T

Bolt size +

(a) Fig. 10.6. External bolting chair.

(b)

192

Design of Supports for Vertical Vessels

Mx

P Orr

(21 sin 7'2) /

(1 + µ) In

ire

+1

[(1

Bearing plate

Fig. 107. Vessel skirt with external bolting choirs.

larger than the hole in the top plate for ease in erection of the vessel. CALCULATION OF COMPRESSION-PLATE THICKNESS. The maximum load on the compression plate at the top of an external chair occurs on the upwind side of the vertical vessel where the reaction of the bolts produces a compression load. The compression plate may be considered to act as a rectangular plate bounded by the two gusset plates, the skirt, and the outside of the plates. The bolt load may be considered to be a uniformly distributed load acting over a circular area equal to the bolt area. The fact that the compression plate is welded to the skirt and gusset plates as indicated in Fig. 10.7 provides additional rigidity on these sides, which tends to compensate for the lack of support on the fourth side. As an approximation the plate will be considered to act as a plate freely supported on four sides. Timoshenko (107) has developed the relationships for a rectangular plate freely supported on four sides with a concentrated load acting as a uniformly distributed load over 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 the circumferential direction, the maximum bending moments M, and Mx are given by Eqs. 10.38 and 10.39, respectively. era)

P M, = — (1 + 47r

2/ sin — In

1

ire

+1

113 47r (10.38)

µ — 72) 9(10.39)

where M, = maximum bending moment along radial axis, inch-pounds Mx = maximum bending moment along circumferential axis, inch-pounds P = maximum bolt load on upwind side (see Eq. 10.35) pounds A = Poisson's ratio (0.30 for steel) In = natural logarithm a = radial distance from outside of skirt to bolt circle, inches / = radial distance from outside of skirt to outer edge of compression plate, inches = gusset spacing, inches e = radius, of action of concentrated load, inches = one-half distance across flats of bolting nut, inches 71, 72 = constants from Table 10.6 A comparison of Eqs. 10.38 and 10.39 using the constants in Table 10.6 indicates that for (b/1) = unity, Mx = Mu, and that for all cases in which (b/1) is greater than unity, M, is greater than Mx and M„ is therefore controlling. After the determination of the size of the bolt and the width of the bearing plate and after the selection of the bolt-circle diameter and gusset spacing, the dimensions a,b,e, and 1 are fixed. The constants - -y1 and 72 may be evaluated by use of Table 10.6, and the maximum bending moments in the radial and circumferential directions may be computed by use of Eqs. 10.38 and 10.39. For the case in which a is selected to be 1/2 and M y is controlling, Eq. 10.38 reduces to: — [(1 4ir

2/ In — + (1 — -y i )

(10.40)

To determine the maximum stress in the compression ring a strip of unit width is considered. For this case, 6M

fmax = 1 21

52

Table 10.6. Constants for Moment Calculation in Compression Ring (107) (Courtesy of McGraw-Hill Book Co.) 2.0 1.2 1.4 1.6 1.8 b/I 1.0 71 0.565 0.350 0.211 0.125 0.073 0.042 72 0.135 0.115 0.085 0.057 •0.037 0.023 0 Note: for a b/I less than 1.0 invert b/I and rotate axes 90°.

Skirt Supports for Vertical Vessels

Or. if fx is assumed to be fau,, (10.41) is = N/(6-Hvilanow.) where / 5 =-- thickness of compression plate, inches fallow, = allowable working stress, pounds per square inch 10.1e Example Calculation 10,2, External-chair Design. An external chair will be designed for a column 8 ft, 0 in. in diameter having 12 bolts 1% in. in diameter with a calculated induced stress of 17,500 psi. The holt-circle diameter is 8 ft, 6 in., and the outside diameter of the bearing plate is 9 ft, 0 in. The gusset height, h, is 12 in. By Fig. 10.6, 1.16

= (9.0(134

= 0.515 in.

A = 9 in. + (41

• Use %-in. plate.)

= 1034 in.,

b = 8 in. + (131 in.) = 93Z in.

By Table 10.4, root area of bolt, Ab = 1.405 sq in. / The bolt load by Eq. 10.35 is: P = fs Ab = (17,500)(1.405) = 24,600 lb By Fig. 10.6, —

(8 ft, 6 in.) — (8 ft, 0 in.) --= 3 in. 2

1=

(9 ft, 0 in.) — (8 ft, 0 in.) — 6 in. 2

t

193

as the upper plate of the bolting ring. Such a continuous ring is preferred when the spacing of external chairs becomes so small that the compression plates approach a continuous ring. As in the case of the compression plate the maximum load on a continuous compression ring occurs on the upwind side of the vertical vessel where the reaction of the bolts produces a compression load on the ring. This load produces a bending stress in the compresson ring. As in the case of external chairs the vertical gusset plates transfer this compression load to the hearing plate. In determining the thickness of the continuous compression ring the assumption is made that each section of the ring between gussets acts as a rectangular plate bounded by the two gusset plates, the shell, and the outer ring. The bolt load will be considered to be a uniformly distributed load acting over the area of the bolt. Therefore the method used in determining the thickness of the compression plates for external bolting chairs is applicable. This method involves the use of Eq. 10.38, 10.39, or 10.40 and of Table 10.6 plus Eq. 10.41. CALCULATION OF GUSSET-PLATE THICKNESS FOR COMPRESSION RINGS. If the gussets are evenly spaced alternately between bolts; the gusset plate may be considered to react as a vertical column. Normally the gusset is welded to the shell, but no credit is taken for the stiffening effect produced by the shell. The moment of inertia>of the gusset about the axis having the least radius3of gyration- is given in 'Appendix J, item I as: i 4 ,)

).

0 in

I—

116 3 12

r2

14r 2

From Table 10.4, e=

nut dimension across flats 2

2.375

2

The compression-plate thickness W: li b 1

9.5

6

L

4.7r

(1Iµ) ln

L

in pi

I it ft

0.134 21 elf

Shell

cif it.

q 111

Strap

+ 1 — yi 11

24,600 1.3 in ( (2) (6) ) ,17r

in

----- 1.58

Substituting in Eq. 10.40 gives: 11,4 =

ri

TA

Interpolating from Table 10.6 giveis: 11 -yi =

1 83

/4.188

Fri Compression ring

+ 1 — 0.11111

0

Guset

= +8200 in-lb

1

I Bea_riing dilate

Substituting into Eq. 10.41 withfaii,;= 17,500 psi Wes: /5 — •

Y— fallow.

\I(6)(8200) = 17,500

1.672

1

Therefore use 1%-in. plate for compression plate. 10.1f Continuous-compression-ring Thickness. Figure 10.8 shows a sketch of a continuous compression ring used

Fig. 10.8.

Skirt with continuous compression ring and strap.

z:

194

Design of Supports for Vertical Vessels

or

/62 12

(10.42)

where a = area of cross section, square inches = radius of gyration, inches 16 = gusset-plate thickness, inches 1 = width of gusset, inches

(Y)x—o =

Equation 4.21 may be used to express the relationship for steel columns in which the value of (h/r) is from 60 to 200. P

f• a — 1

force, Q0 (see (Fig. 6.3) produced in shells with closures, and the calculation may be treated accordingly. In Chapter 6 the following relationships were derived:

18,000 (h2/18,000r2)

(4.21)

(6.76) (61 — 201, (25'Mo Qo) D1 0 In detail a of Fig. 10.6 y is the horizontal deflection of the skirt corresponding to yi in Fig. 6.3 and varies with distanc,e above the compression ring in accordance with Eq. 6.65. Applying the boundary condition that the slope of the deflection curve dy / dx is equal to zero at the top of the ring where x = 0, we obtain:

((Adx),o = 0 = 202D1

where h = height of gusset, inches Substituting Eq. 10.42 into this relationship gives.

fallow. — 1

18,000 (h2/1500162)

(10.43)

Bolt load P a 116

(10.44)

(Y)x—o — 215,3D1 (ftMo Qo)

(10.49)

Qo = 403D1y

(10.50)

M0 = —202D1y

(10.51)

4 13(1 — ,l12)

.4

/

(6.86)

r 212

Or

h2(bolt load) — 0 (10.45) 1500

Examination of Eq. 10.45 indicates that when the gusset height, h, is small, the third term in the equation may be disregarded. In this case Eq. 10.45 reduces to the relationship for straight compression without column action or bolt load (10.46) 16 18,000/ Equations 10.45 and 10.46 are based on the asumption that the compression plate is sufficiently thick for the bolt load to be transferred to the gusset plates without introduction of eccentric action. The stiffening resulting from the welding of the compression plates and gussets to the shell introduces a margin of safety which justifies the above assumption. If the gussets are not evenly spaced, an eccentric loading will result in an induced bending moment. The thickness of such gussets may be proportioned empirically, as in the ease of gussets for external chairs. 16 = 116

(10.48)

where

18,000 1 + (h 2/150016 2)

18,000/4 3 — (bolt load)/62 —

Qo)

Solving Eqs. 10.48 and 10.49 for M0 and Qo gives:

Substituting in Eq. 10.43 gives: Bolt load 116

(2$Mo

Noting that yi of Fig. 6.3 is taken as equal to —y in Fig. 10.6, detail a, we obtain:

The allowable stress, Allow., in Eq. 10.43 must be:

fallow. —

(6.75)

(0Mo + Qo) 2/33D1

(10.47)

10.1g Reaction of External Bolting Chairs and Compression Rings. The use of external bolting chairs or a compression ring results in a loading condition that produces a reaction in the skirt. This reaction, R, is similar to the shear

DI —

.E13 12(1 — u')_.--

(6.15)

By Eq. 6.84 w.-=lE r2 y

(10.52)

where w = load, pounds per linear inch = skirt thickness, inches r = radius of curvature, inches Substituting Eqs. 6.15 and 10.52 into Eqs. 10.50 and into Eq. 10.51 gives: •

Qo =

$312....2 wr

(10.53)

=

3(1 — _ 02i2u,r2 Mo — 6(1 —

—w 2132

(10.54)

Taking a summation of moments about the junction of the skirt with the bearing plate gives: P(a) = Qohm where (20 is the force per linear inch on the skirt and is assumed to act over an arc distance of m, or Pa Qo = — mh

(10.55)

Skirt Supports for Vertical Vessels Substituting Eq. 10.53 into 10.55 gives: Paj mh

(10.56)

This thickness can be reduced by increasing the gusset height. Assuming a gusset height of 18 in. rather than 12 in. will reduce the skirt thickness to: t = 1.214(+1)" = 0.926 in. or 1.0 in.

Substituting Eq. 10.56 into Eq. 10.54 for w gives:

fl3t2Par2

M — D

(10.57)

6(1 — u2) mh

For a strip of unit width under flexure j 3Par2 (1 _ to)inh

6M0 12

(10.58)

Substituting for /3 by Eq. 6.86 and solving for t gives: [3(1

10,1”

— u2)l f2[ LIL

(1 —

AI)"

r

mhf

For steel in which p = 0.3, t = 1.76

( Pa

)35

195

10.11 Thermal Stresses in the Skirt. For the case in which the vessel is operated at a temperature considerably different from atmospheric temperature, a thermal stress may be induced in the skirt as a result of the temperature gradient near the junction of the skirt and the vessel. TEMPERATURE GRADIENT IN SKIRT. To minimize the temperature gradient in the skirt, the skirt may be insulated both inside and out. Skirts of vessels are insulated inside and out for fire protection when manways are cut into the skirt. The modulus of elasticity decreases rapidly with increasing temperature above 600° F with resulting loss in elastic stability. F. E. Wolosewick (160) has given an approximate equation for the skirts of vessels with 2 to 4% in, of insulation both inside and out.

(10.59)

mhisnow.

T x = (Tv — 50°) — 6.037x — 0.289x2

where i = skirt thickness required to resist reaction of external chairs or compression ring, inches r = radius of skirt, inches m = 2A (see Fig. 10.6) or bs (bolt spacing) P = maximum bolt load, pounds a = radial distance from outside of skirt, to bolt circle, inches h = gusset height, inches 10.1h Example Calculation 10.3, Reaction of a Bolting Ring. The tower described in Example Calculation 10.1 is to be modified so that it has a bearing plate extending 6% in. out from the skirt with a bolt circle 3% in. outside the skirt. Twenty-four bolts 2) in. in diameter are to be used with a continuous compression ring, and the gusset height is to be 12 in. Determine the required thickness of the skirt to resist the reaction of the bolting ring. The maximum induced stress in the bolts is 17,450 psi, and the maximum allowable stress in the skirt is 20,000 psi.

0.009x3 0.00007x4 (10.60) Differentiating with respect to x gives: dTx dx

—6.037 — 0.578x + 0.027x2 — 0.00028x3

(10.61)

where Tx = temperature of skirt at x distance below junction•of skirt and shell, degrees Fahrenheit Tn = temperature of fluid in vessel bottom. degrees Fahrenheit x = distance below junction of skirt and shell, inches

hut.. = 20,000 psi

THERMAL EXPANSION. As an insulated vessel is brought up to operating temperature, it will undergo thermal expansion. If there is no restraint to this expansion, no stress will be induced. The metal both in the skirt and in the shell at the junction will have the same temperature. From the junction to the foundation a temperature gradient will exist, which will tend to produce a varying thermal expansion. At any given point in the skirt the radial thermal expansion, y, is proportional to the coefficient of thermal expansion, a, the radius of the skirt, r. and the temperature difference Tl, or

h= 12 in.

y = ar(T,, — Tx) = arT'

a = 3%-in. m=

r(126.5) — 16.6 in. 24

p = 17,450 X 3.72 sq in. per bolt r = 60 in. By Eq. 10.59 t = 1.76 (

Pa )" mhiaiiow.

Substituting gives:

450) (3 72) (3 25)] . ,33

t = 1.76 [ (17 - ' (16.6)(12)(20,000) = 1.214 in. or 13& in.

(10.62)

where y = radial thermal expansion, inches a = coefficient of thermal expansion inches per inch per degree Fahrenheit r = radius of skirt, inches TI = (T,, — Tx) = temperature of vessel bottom minus skirt temperature, degrees Fahrenheit Differentiating Eq. 10.62 with respect to x. the distance along the skirt from its junction with the vessel, gives:

(60)

dy _ ar dT 1 dx dx

196

Design of Supports for Vertical Vessels

but dTt = d(T, — Tx ) = — dTx ; therefore

and

dy _ ar dTx dx dx

drx I Qo = a di32 Di —

(10.63)

The term dy/dx represent the slope of the skirt from the vertical as a result of thermal deformation. This deflection can be compared to the deflection dyi/dx for a cylindrical shell joined to a flat-plate closure, shown in Fig. 6.3. At the junction, where x = 0, STRESS FROM BENDING MOMENTS AND SHEAR.

A comparison of Eqs. 10.69 and 10.70 indicates that

(6.75)

Cy)

—

AXIAL THERMAL STRESS, fat , AT JUNCTION WITH SHELL.

(See Eq. 6.122.)

1 20 3 D1

ar dTI (dy dx dx x= o

(riMo

1 2132D,

/2

I

CIRCUMFERENTIAL THERMAL STRESS,

SAM° ± Qo)

I

LE —

Lt =

(10.64)

(202D1 ) :=-- faro

(10.73)

iL6MO 12

(10.74)

3 Qo Ia di32 D1 dTx 21 dx

T

(10.75)

a = 7.6 X 10-6 deg F dT 1) dx )

2ar#D1 (T's

E = 25.5 X 106 psi

(10.65)

= 0.27

dx ) —

Substituting into Eq. 10.64 gives: dT dx

andT i(2

(3 2 Di) = 4ar62D i (T1/3

Qo

Qo = 2ar/32D1 (2T1(3 — d T) dx

(10.66)

At the junction of the skirt and bottom dished head T' =T2, — Tx , and T„ = Tx ; therefore T' = 0. And, therefore, dT1 Mo = a dO D1 —

(10.67)

dx

—

/12)

r2/2

=

— 0.272)

— 0.227

(65)2(0.5)2

By Eq. 6.15 D1 —

therefore

Eta 12(1 — p.2)

25.5 X 106(0.5)3

— 28.7 X 10 4

12(1 — 0.272)

The temperature gradient at the junction, x = 0, by Eq. 10.61 is: dTx

= —6.037

Axial thermal stress:

Substituting into Eq. 10.69 gives:

and Qo = —a 02 Di

dT dx

(10.68)

Or

—a d(3 D 1

(from Fig. 8.6)

By Eq. 6.86

20.1.10 = 4arO2D1 (T1/3

Mo

/2

10.1j Example Calculation 10.4, Thermal Stresses. Consider a vessel having a diameter of 130 in. and a skirt thickness of in. insulated inside and out with the skirt suppLting a shell in which the bottom temperature is 700° F. Assume that the temperature distribution in the skirt is given by Eq. 10.60. Calculate the thermal stresses at the junction.

therefore Mo

//WO

(See Eq. 6.121.)

Adding the two equations gives:

dx

JUNCTION

SHEAR THERMAL STRESS, .fst, AT JUNCTION WITH SHELL.

(2,6Mo + Qo)

arT 1(2,8 3.D 1) = —13Mo — Qo

ar dr!.

AT

But by Eq. 10.71 0M0 = —Qo; therefore

Clearing fractions on the right side of these equations gives:

arT1(2133M)

+

t

Qo)

ar dT t (2/32D1) = 20M0 Qo dx

Li,

(See Eq. 6.125.)

fee —

Substituting Eq. 10.62 into Eq. 6.75 and Eq. 10.63 into Eq. 6.76 gives:

—6a c/6 DI 0; I (10.72) 12 dx I

Moc — 6M0

(6.76)

(20Mo Qo)

202Di

arT1 = yx—o

fat =

WITH SHELL. '

1

(10.71)

i3Mo = — Qo

i

3 Di (i3Mo Qo) 21

(10.70)

. dx

.

dT x

dx

Mo = — dp

dTx dx

— (7.6 X 10-6)(130)(0.227)(28.7 X 104)(-6.037) (10.69)

= 388 in-lb per in.

Lug Supports for Vertical Vessels By Eq. 10.72 fat =

6(388) (0.5)2 — 9320 psi

6/1/0 j2

Circumferential thermal stress: By Eq. 10.70 Q0 = a d132

are setup on the windward side when the vessel is empty Ice-ea—Use in this case the dead load—is—subtracted-from., tbe Wind-100, Therefore the stresses on the leeward side are the determining factor for design of the supports. The maximum total compression load in P pounds in the most remote column is (164):

D dT, dx

P=

= (7.6 X 10-6)(130)(0.227)2(28.7 X 10')(-6.037) -- —88.2 lb per in. By Eq. 10.74 fet =- ,

a6/1/0 — R0.27)(9320)] = 2115 psi 12

I Shear thermal stress: By Eq. 10.75 fst

s Q,

=

—88-? — 264.6 psi

0.5

10.2 LUG SUPPORTS FOR VERTICAL VESSELS The choice of the type of supports for a vertical pressure vessel depends on the available floor space, the convenience of location of the vessel according to operating variables, the size of the vessel, the operating temperature and pressure, and the materials of construction. Brackets or lugs offer many advantages over other types of supports. They are inexpensive, can absorb diametrical expansions by sliding over greased or bronze plates, are easily attached to the vessel by minimum amounts of welding, and are easily leveled or shimmed in the field. As a result of the eccentricity of this type of support, compressive, tensile, and shear stresses are induced in the wall of the vessel. The tensile and compressive forces cause indeterminate flexural stresses which must be combined with pressure stresses circumferentially and longitudinally. The shear forces act in a direction parallel to the longitudinal axis of the vessel, and the shear stress induced by these forces is relatively so small that they are often disregarded. Lug supports are ideal for thick-walled vessels since the thick wall has a considerable moment of inertia and is therefore capable of absorbing the flexural stresses due to the eccentricity of the loads. In thin-walled vessels, however, this type of support is not convenient unless the proper reinforcements are used or many lugs are welded to the vessel. If a vessel with lug supports is located out of doors the wind load, as well as the dead-weight load should be considered in the calculation of P. However, as lug-supported vessels are usually of much smaller height, than skirt-supported vessels, the wind loads may be a minor consideration. The wind load tends to overturn the vessel, particularly when the vessel is empty. The weight of the vessel when filled with liquid tends to stabilize it. The highest compressive stresses in the smpslrts occur on the leeward side when the vessel is full because dead Jed and wind load are additive.„,„ The highest tensile stresses

197

4P,„(H L) I IV riDoc

(10.76)

where P,„ = total wind load on exposed surface, pounds H= height of vessel above foundation, feet L vessel clearance from foundation to vessel bottom, feet Dbc diameter of anchor-bolt circle, feet n number of supports Td' = weight of empty vessel plus weight of liquid and other dead load, pounds 10.2a Lugs with Horizontal Plates. Figure 10.9 shows a sketch of a vessel supported on four lugs, each lug having two horizontal-plate stiffeners. Such lugs are of essentially the same design as that shown in Fig. 10.6 for external chairs, and the same design procedure may be used. This type of lug uses to advantage the axial stiffness and strength of the cylindrical shell to absorb the bending stresses produced by the concentrated loads of the supports. Both the top and bottom plates should have continuous welds as the maximum compressive and tensile stress occurs in these two plates, respectively. These welds and the intermittent welds of the vertical gussets to the shell carry the vertical shear load. The load, P, on the column has a lever arm, a, measured to the center line of the shell plate. This moment

r-

Qo 1

h

1 ••••••-3.-1 0

H

Y

Fig. 10.9. Sketch of vessel on four-lug supports with horizontal-plate stiffeners.