Pressure Vessel Handbook
Pressure Vessel Handbook
Pressure Vessel Handbook
HANDBOOK
Twelfth Edition
with foreword by
PaulButhod
Professor of Chemical Engineering
University of Tulsa
Tulsa, Oklahoma
Eugene F. Megyesy
The design procedures and formulas of the ASME Code for Pressure
Vessels, Section Vlll Division I have been utilized as well as those
generally accepted sources which are not covered by this Code. From
among the alternative construction methods described by the Code the
author has selected those which are most frequently used in practice.
A large part of this book was taken from the works of others, with some
of the material placed in different arrangement, and some unchanged.
The author wishes also to thank all those who helped to improve this
new edition by their suggestions and corrections.
· SuggestioQs and criticism concerning some errors which may remain
in spite of all precautions shall be greatly appreciated. They contribute to
the further improvement of this Handbook.
Eugene F. Megyesy
7
PART I.
b. Lingitudinal The
. smaller of Sa or the value of
compressive stress factor B determined by the procedure
described in Code UG 23 (b) (2)
c. General primary membrane stress
induced by any combination of
loadings. Primary membrane stress l.SSa
plus primary bending stress induced S =(see above)
by combination of loadings, except a
d. General primary membrane stress -1.2 times the stress permitted in a., b.,
IN REFERENCES THROUGHOUT THIS BOOK "CODE" STANDS FOR ASME induced by combination of earth- or c. This rule applicable to stresses
BOILER AND PRESSURE VESSEL CODE SECTION VIII DIVISION I _ AN quake or wind pressure with other exerted by internal or external pressure
AMERICANSTANDARD. ' loadings. Seismic force and wind or axial compressive load on a cylinder.
2001 EDITION pressure need not be considered to
act simulta neously.
14 15
INTERNAL PRESSURE
STRESSES IN CYLINDRICAL SHELL
I. OPERATING PRESSURE
The pressure which is required for the process, served by the vesse I, at which
the vessel is normally operated.
Unifonn internal or external pressure induces in the longitudinal seam two times larger unit 2. DESIGNPRESSURE
stress than in the circumferential seam because of the geometry of the cylinder.
The pressure used in the design of a vessel. It is recommended to design a
vessel and its parts for a higher pressure than the operating pressure. A
A vessel under external pressure, when other forces (wind, earthq11ake, etc.) are not
design pressure higher than the operating pressure with 30 psi or 10 percent,.
factors, must be designed to resist the circumferential buckling only. The Code
whichever is the greater, will satisfy this requirement. The pressure of the
provides the method of design to meet this requirement. When other loadings are
fluid and other contents of the vessel should also be taken into consideration.
present, these combined loadings may govern and heavier plate will be required
See tables on page 29 for pressure of fluid.
than the plate which was satisfactory to resist the circumferential buckling only.
3. MAXIMUM ALLOWABLE WORKING PRESSURE
The compressive stress due to external pressure and tensile stress due to internal pressure
shall be detennined by the fonnulas: The internal pressure at which the weakest element of the vessel is loaded
to the ultimate permissible point, when the vessel is assumed to be:
FORMULAS (a) in corroded condition
CIRCUMFERENTIAL LONGITUDINAL (b) under the effect of a designated temperature
JOINT JOINT (c) in normal operating position at the top
(~under the effect of other loadings (wind load, external pressure, hydro-
PD static pressure, etc.) which are additive to the internal pressure.
S2=-
2t When calculations are not made, the design pressure may be used as the
maximum allowable working pressure (MA WP) code 3-2.
Primary Service
Pressure Rating 150 lb 300 lb 400lb 600lb 900lb 1500 lb 2500lb
Hydrostatic SheD Test
Pressure 425 1100 1450 2175 3250 5400 9000
A Pneumatic test may be used in lieu of a hydrostatic test per Code UG-1 00
6. JOINT EFFICIENCY
The efficiency of different types of welded joints are given in table on page
172. The efficiency of seamless heads is tabulated on page 176.
The following pages contain formulas used to compute the required wall
thickness and the maximum allowable working pressure for the most
frequently used types of shell and head. The formulas of cylindrical shell are
given for the longitudinal seam, since usually this governs.
The stress in the girth seam will govern only when the circumferential joint
efficiency is less than one-half the longitudinal joint efficiency, or when
besides the internal pressure additional loadings (wind load, reaction of
saddles) are causing longitudinal bending or tension. The reason· for it is
that the stress arising in the girth seam pound per square inch is one-half of
the stress in the longitudinal seam.
1
PR p = 2SEt
= 2SE + 0.4P R- 0.41
See notation on page 22.
18 19
r - 1 1- - - - - - - ·. . .- - - - - - - - - - - - - - - _ _ : : :
i
INTERNAL PRESSURE EXAMPLES
/
i
/
t
t
PR P= SE_t_ I Determine the required thickness, Determine the maximum allowable
working pressure P for o.500 in. thick
~~ SE-0.6P R+0.6t J tofashell shell when the vessel is in new condition.
R
+-·-t-· ·~ / 100 X 48.125 .
~\ WJ I. Usual!~ the stress in the long seam is governing. See
t=2(f,'OOO X0.85-0.6X100 =0.2S4 m.
0.125 in.
P=20,000X0.85X0.500 _ .
48 + 0.6 X 0.500 - 176 psi
prccedmg page. +CA.
2. Wh;n the wall thickness exceeds one half of the inside 0.409in.
radiUS or P exceeds 0.385 SE, the formulas given in
the Code Appendix 1-2 shall be applied. Use 0.500 in. plate
B SEEDESJGNDATAABOVE SEEDESIGNDATAABOVE
SPHERE and HEMISPHERICAL HEAD
The head furnished without straight
flange.
fb PR
t=2SE-0.2P
P= 2SEt
R+0.2t
Determine the required thickness
t of a hemispherical head. '
Determine the maximum allowable
working pressure, P for 0.3125 in. thick
head, when it is in new condition.
~~~
.,. 1
100X48.125 =0 142 .
2X20,000X0.85-0.2X fOO . m. p 2X20,000X0.85X0.3125 _ .
i R f
I. For heads without a straight flange, use the efficiency
of the head to shell joint if it less than the efficiency
of the seams in the head. +C. A. 0.125in.
48+0.2X0.3125 -
221
pst
0.267in.
2. When the wall thickness exceeds 0.356 R or P exceeds
0.665 SE. the formulas given in the Code Appendix Use 0.3125 in. plate
l-3, shall be applied.
c 2:1 ELLIPSOIDAL HEAD ~
SEE DESIGN DATA ABOVE SEE DESIGN DATA ABOVE
Determine the required thickness of a
seamless ellipsoidal head.
Determine the maximum allowable
PD 2SEt
·s-~
t lOO X'9625 ' · working pressure, P for 0.250 in. thick
2SE-0.2P P=
D+0.2t 1
2X20,000X 1.0~0.2X100 4
=0.2 lin.
seamless head, when it is in corroded
condition.
~
PR SEJ 100 48 working pressure, P for 0.4375 in. thick
t X 0.283 in
,_ SE + 0.4P p- R - 0.4t 20,000 X 0.85-0.4 XIOO shell when the vessel is in new condi-
tion.
0.125 in.
+CA. 0.408 in. P= 20,000 X 0.85 X 0.4375 psi
155
1. Usually the stress in the long seam is governing. See
48-0.4 X 0.4375
page 14 Use: 0.4375 in. thick plate
2. When the wall thickness exceeds one half of the inside
radius or P exceeds 0.38S SE, the formulas givenf.in
the Code Appendix 1-2 shall be applied.
SEE DESIGN DATA ABOVE SEE DESIGN DATA ABOVE
B
SPHERE and HEMISPHERICAL HEAD Head furnished without straight flange.
I Determine the required thickness, t of a
~
Determine the maximum allowable
hemispherical head.
t - 2SE
PR
+ 0.8P
p 2SEt
- R -0.81 i
l
100X48
t 2X20,000X0.85+0.8X100
141
°·
in.
working pressure, P for 0.3125 in. thick
head, when the vessel is in new
condition.
t
~
rt +CA. 0.125in. P=2X20,000X0.85X0.3125 .
I. For heads without a straight flange, use the efficiency 48-0.8 X0.3125 222 pst
of the head to shell joint if it is less than the efficiency 0266in.
of the seams in the head. Use: 0.3125 in. min. thick head
2. When the wall thickness exceeds 0.3S6 R or P exceeds
0.66S SE, the formulas given in the Code Appendix
1-3, shall be applied.
SEE DESIGN DATA ABOVE SEE DESIGN DATA ABOVE
c 2: 1 ELLIPSOIDAL HEAD
Determine the required thickness t of a Determine the maximum allowable
seamless ellips()idal head. working pressure, P for 0.375 in. thick
h~
PD p 2SEt
t 100X96 head, when it is in new condition.
---·
2SE+ 1.8P D· -1.8t
t 2X20,000X1.0+1.8X100 °'239 in.
1. D f I. For ellipsoidal heads, where the ratio of the major and
minor axis is other than 2: I, see Code Appendix l-4(c).
+CA.
h = D/4
24 ·,_ : 25
r
INTERNAL PRESSURE EXAMPLES
FORMULAS IN TERMS OF OUTSIDE DIMENSIONS
DESIGN DATA: heads
NOTATION P = I 00 psi design pressure R 48 inches outside radius
=
D = Outside diameter, inches ' S = 20,000 psi stress value of D 96 inches outside diameter
P = Design pressure or max. allowable a = One half of the included (apex) a.= 30"onehalfoftheapexangle
working pressure psi angle, degrees SA 515-70plate@5000F
S = Stress value of material psi, page L = Outside radius of dish, inches E = 0.85, efficiency ofspot-examined L 96 inches outside radius of dish
189 r = Inside knuckle rlldius, inches joints t Required wall thickness, inches
E = Joint efficiency, page 172
R = Outside radius, inches
t = Wall thickness, inches ·
C.A. =-Corrosion allowance, inches
' l E 1.00, joint efficiency ofseamless C.A. = 0.125 inches corrosion allowance
: SEEDESIGNDATAABOVE SEEDESIGNDATAABOVE
D cos 30° = 0.866
CONE AND CONICAL SECTION Determine the maximum allowable
Determine the required thickness, t working pressure, P for 0.500 in. thick
of a cone cone in new condition .
...L ' 100X96 .
I l ~
PD P= 2SEtcosa 326
t r-2X0.866X(20,000X0.85+{).4X1 00)=0 m.
2 cos a (SE +0.4P) D -0.8tcos a
P=2X20,000X0.85X0.500X0.866 .
c:::: ::::::::1 +C.A. 0.125 in. 96 -(0.85 X 0.500 X 0.866) 153 psi
JR
\
~-
t=0.5625 -0.125 =0.4375
0.88SPL SEt
I t P= 0.88SL-0.8t 20,000X 1.0X0.4375
i ~
SE+0.8P +C. A. 0.125 in. p 103psi
0.885 X96-0.8 X0.4375
!.. ~- \1
0.548in.
D When Ltr Less Than 16 213 Use: 0.5625 in. min. thick head
f:
L/r
1.00
1.25
1.50
1.75
2.00
VA LUES OF FACTOR M
2.25
2.50
2.75
3.00
3.25
3.50
4.00
4.50
5.00
5.50
6.00
6.50 II i ~ head.
~
•· , · - .
100X96X 1.75 .
0 419
t=2X20,000X 1.0+ f00(1.75-02) · m.
. thick seamless head when the vessel is
in corroded condition.
M 1.00 1.06
1.08
1.10 1.15
1.17
1.18 1.22
1.25
1.28 1.34
1.36
1.39 I ' 2 X 20,000 X I.0 X 0.4375 .
! i
11.03 1.13 1.20 1.31 +C. A. 0.125 in.
l p 1.75 X96-0.4375(1.75 -0.2)- 104 psi
Ltr
7.00 8.00 9.00 10.0 11.0 12.0 14.0 16.0 • ,, 0.544in.
17.~0 8.50 9.50 10.5 11.5 13.0 u.o t6t t Use 0.5625 in. min. thick head
'
M
1.41
44 Itit 48
1.46
1.52 ! 1.56
1.50 1.54 1.58
1.60
1.62
lt.65
1.69
11.72
1.75
It 77 ~I NOTE: When the ratio of Llr is greater than 16~ , (non-Code construction) the values of
• THE MAXIMUM ALLOWED RATIO : L • t =D (see note on facing page) ~ M may be calculated by the formula: M = Y. (3 + Wr)
'
26 27
il'l
~~ of the vessel and is due to the height of the fluid above the point at which the
Temperature, F MAXIMUM ALLOWABLE NON-SHOCK PRE,SSURE PSIG. @ [, pressure is considered.
-20 to 100 285 740 990 1,480 2,220 3,705 6,170 !i: The static head when applicable shall be added to the design pressure of the
200
300
260
230
675
655
900
875
1,350
1,315
2,025
1,970
3,375
3,280
5,625
5,470 ;
!f:
'
i vessel.
The tables below when applicable shall be added to the design pressure of the
m
400 200 635 845 1,270 1,900 3,170 5,280 ' ~ water.
~
u; To find the pressure for any other fluids than water, the given in the tables shall
be be multiplied with the specific gravity of the fluid in consideration.
500
600
170
140
600
550
800
730
1,200
1,095
1,795
1,640
2,995
2,735
4,990
4,560
l~·
Pressure in Pounds per Square Inch for Different Heads of Water
650 125 535 715 1,075 1,610 2,685 4,475 Head
Feet 0 I 2 3 4 5 6 7 8 9
700 110 535 710 1,065 1,600 2,665 4,440
0 0.43 0.87 1.30 1.73 2.16 2.60 3.03 3.46 3.90
10 4.33 4.76 520 5.63 6.06 6.49 6.93 7.36 7.79 8.23
750 95 505 670 1,010 1,510 2,520 4,200 20 8.66 9.09 9.53 9.96 10.39 10.82 11.26 11.69 12.12 12.56
800 80 410 550 825 1,235 2,060 3,430 30 12.99 13.42 13.86 1429 14.72 15.15 15.59 16.02 16.45 16.89
i
850 65 270 355 535 805 1,340 2,230 40 17.32 17.75 18.19 18.62 19.05 19.48 19.92 20.35 20.78 2122
50 21.65 22.08 22.52 22.95 23.38 23.81 2425 24.68 25.11 25.55
900 50 170 230 345 515 860 1,430 00 25.98 26.41 26.85 2728 27.71 28.14 28.58 29.01 29.44 29.88
70 30.31 30.74 31.18 31.61 32.04 32.47 32.91 33.34 33.77. 34.21
950 35 105 140 205 310 515 860 ;: 80 34.64 35.07 35.51 35.94 36.37 36.80 37.24 37.67 38.10 38.54
1 000 20 50 70 105 155 260 430 .,i SQ 38.97 39.40 39.84 4021 40.70 41.13 41.57 42.00 42.43 42.87
' 1 NOTE: One foot of water at 62° Fahrenheit equals .433 pound pressure per square
inch. To find the pressure per square inch for any feet head not given in the table
Ratings apply to NPS ~ trough NPS 24 and to materials: above, multiply the feet times .433.
A 105 (1) A 350 Gr. LF2 (1) A 350 Gr. LF6 Cl. 1 (4) A 216 Gr. WCB (1) Heads of Water in Feet Corresponding to
A515Gr. 70(1)A516Gr. 70(1)(2)A537Cl. 1 (3) i Certain Pressure in Pounds per Square Inch
NOTES: ; Pres-
(1) Permissible, but not recommended for prolonged use above 800 °F. j sure, 0 I 2 3 4 5 6 7 8 9
Lbs.
(2) Not to be used over 850 °F.
0 2.3 4.6 6.9 9.2 11.5 13.9 16.2 18.5 20.8
(3) Not to be used over 700 °F. 10 23.1 25.4 27.7 30.0 32.3 34.6 36.9 39.3 41.6 43.9
(4) Not to be used over 500 °F. 20 46.2 48.5 50.8 53.1 55.4 57.7 60.0 62.4 64.7 67.0
Flanges of ANSI B 16.5 shall not be used for higher ratings except where it 30 69.3 71.6 73.9 76.2 78.5 80.8 83.1 85.4 87.8 90.1
40 92.4 . 94.7 .9'7..0 99.3 101.6 103.9 106.2 108.5 110.8 113.2
is justified by the design methods of the Code. 50 115.5 117.8' 120.1 . 122.4 124.7 127.0 129.3 131.6 133.9 136.3
Ratings are maximum allowable non-shock working pressures expressed 00 138.6 140.9 143.2 145.5 147.8 150.1 152.4 154.7 157.0 159.3
as gage pressure, at the tabulated temperatures and may be interpolated 70 161.7 164.0 f66.3 168.6 170.9 173.2 175.5 177.8 180.1 182.4
80 184.8 187.1 189.4 191.7 194.0 196.3 198.6 200.9 203.2 205.5
between temperatures shown. I 00 207.9 210.2 212.5 214.8 217.1 219.4 221.7 224.0 226.3 228.6
Temperatures are those on the inside of the pressure-containing shell of the
'· NOTE:~~r~~~U~d ofpres~ure per square inch of water equals 2.309 feet of water
flange. In general, it is the same as that of the contained material. at 62° Fa t. Therefore, to find the feet head of water for any pressure not
Flanged fittings shall be hydrostatically tested. given in the table above, multipy the pressure pounds per square inch by 2.309.
30 31
TABLES
For quick comparison of required plate thickness and weight for various
materials and at a different degree of radiographic examination.
EXTERNAL PRESSURE
.A .Stress.yalues at temperature -20° to 500 °F. DESIGN PRESSURE
(
SA 53 B When Code Symbol is to be applied, the vessel shall be designed and
' stamped with the maximum allowable external working pressure. It is
SA285.C SA 515-60 ·SA 515-70
SA 516-60 SA 516-70 recommended that a suitable margin is provided when establishing the
85% J. E. 13,345 14,535 17,000 maximum allowable external pressure to allow for pressure variation in
100% J. E.
= 15,700
~-12_6_00~x~1._08_=~13~6_08~l_b_.__________________________~t
i
'
32 33
VESSEL
When the value of D0 1t is less than 10, the
formulas given in the Code UG-28(c)(2) shall
. -..,.
N See page 40 for design of stiffening rings.
WITH STIFFENING RING be applied.
34 35
r ! following thicknesses.
(1) The thickness as computed by the formulas
A= 0.125/(86.4/0.3125) = 0.00045
B = 6100 from chart (page 43 ),P" - B/(R 0 1 t)l= 6100/276 = 22.1 psi.
+
given for internal pressure using a design pres-
sure 1.67 times the external pressure and joint Since the maximum allowable pressure P" is greater than the design pressure
efficiency £ =1.00. P the assumed thickness is satisfactory.
D,
(2) The thickness proofed by formula P0 = B/R0 /t
whereR.,=0.9 Du, and B to be determined as for
sphere.
SEE DESIGN DATA ABOVE. Procedure (2.)
ASME FLANGED AND DISHED tmAD
(TORISPHERICAL HEAD) Assume a head thickness: t = 0.3125 in., R.=D. = 96 in.
A = 0.125/(96/0.3125) = 0.0004
The required thickness and maximum allowable pres- =
B 5200 from chart (page 43), P... B/(R 0 /t) = 5200/307 = 16.93 psi.
sure shall be computed by the procedures given for Since the maximum allowable pressure P" is greater than the design pressure
ellipsoidal heads. (See above)R 0 maximum=D, P the assumed thickness is satisfactory. ·
36 37
NOTES
EXTERNAL PRESSURE
FORMULAS
""'i:o
II. Using 2 stiffening rings equally
spaced between one-third the
depths of heads (see figure),
t = Minimum required wall thickness of shell, in. Ls= 196in.
I. Select the type of stiffening ring and detennine its cross sectional area A. s::
~
-;- - ____,
III. The moment ofintertia of the
II. Assume the required number of rings and distribute them equally between selected angle: 11.4 in.
jacketed section, cone-to-shell junction, or head line at 11.3 of its depth and ~
detennine dimension, Ls.
s::
~ i:o - ---· - ""'
\0 1. The value of Factor B:
III. Calculate the moment of inertia of the selected ring or the moment of inertia of 00
B =% [PD 0 /(t + AJLs)] =
the ring combined with the shell section (see page 95). t- - - % [15x96/(0.5 + 3.03/196)]
IV. The available moment of inertia of a circumferential stiffening ring shall not be I ""' =2095
less than detennined by one of the following fonnulas: ~ 00
in 2. Since the value of B is less
I' _ D.2Ls (t+A/L)A I _ D.2Ls (t+AjL)A than2500,
s - 10.9 ·' - 14 i
A =2B/E=
The value of A shall be detennined by the following procedure: i: l: 2 X 2095/27,000,000 = 0.00015
1. Calculate factor B using the fonnula:
rv. The required moment of inertia:
B=%[ PD. ]
t+A/Ls
2. Enter the applicable material chart (pages43 -A7) at the value of Band move
horizontally to the curve of design temperature. When the value of B is less than
_ [Da2Ls(t+As!L) A] =961 x 196x(0.5 +3.03 I 196)X 0.00015 = 9 .97 in.4
2500, A can be calculated by the fonnula: A = 2B/E.
Is- . 14 . . 14 --
3. From the intersection point move vertically to the bottom of the chart and read the
value of A. Since the requlred m~ment of inertia (9,97 in. 4) is smaller than the moment of
4. Calculate the required moment of inertia using the fonnulas above. inertia ofthe selected angle (11.4 in. 4) the vessel is adequately stiffened.
If the moment of inertia of the ring or the ring combined with the shell section is greater Stiffening rings may be subject to lateral buckling. This should be considered
than the required moment ofinertia, the stiffening of the shell is satisfactory. Otherwise in addition to the required moment of inertia.
stiffening ring with larger moment of inertia must be selected, or the number of rings
shall be increased. See pages 95-97 for. stiffening ring calculations.
Stiffening ring for jacketed vessel: Code UG-29 (f)
UDO ....,.,. _..,..; N
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[. tl.i. /.~.9~.2~---~.-•-.C-&.!\$C((#)!$~·!i!!$3 40 ::a, WJ$44glJI;:;:;pa;< W oj.,..,,r/k'> oc"'""'-"·""*·""""''"'+"!'A,M""'"""""""'"''"~~~~·
25,000 l=l
I I30,011 _
Q) Q)
..<::tt!fi
...... Q)
up to ~
500 F
20.000 Be'O
"' ~ c
--
18,000
v I I =: "' 0
700 F- 16.000 ~ ~-~
..,......- v I
sop~-
I
...... II)
• 0
Q)
v ...,....,... v
.- 1---' ,....
!...-- ~
~
~ f-"
~ _. 1-
,......
,...v
I-"
12,000
10.000 =
_,
g.St;~
Cde~=
> II) 0 II)
Q. •
9,000
i- .....
lo"4 C.N!-o
? / ..;<' i-
....,., ....
8,000 o
<.....
...2
.....
e ·c0 .so:s
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1) ./ 7,000
r Ql +ooo~ .....r:: '-
U boog_
IF. / 6.000
<
~
.s ..<:: e
..<::
~'+-""ll>
rt. "' ..<:: .....
0
~] -~ .s
/, 'II "' .... Q)
5,000
1// o:sll) '+-
E • 29.0 x 10 6 0 II) ~ 0
~ ...... rt/J. 4,000 ..5 ..<:: -~
E =27.0
E = 22.8
K
E = 24.5 x 106
X
106
106
...... l/.
r.......tt rt
w FIG.CS-2 3,500
.......... ~ .....
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··'+-~o
&-; 0 " ' !-<
"0
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I Ill z .s
2 3456789
rtl'ii bi71V 2 3 4 5 6 7 89 2 3 456789 2 3 4 5 6 789 2.500
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.001 .01 .1
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FACTOR A
<I)
<I)
20.000
500 F
18,000
Be'S
,. ~,..-V I I
700 F ~ "' enfii ot::
.....-~-"' v v I I
16,090
......
<d <d · - ....
- v---
vv v
" <I)
c..>
sop~-
v 1-1- ...- I-"
900 F
14,000
12.000 . 0
......
~-
(1)
,,""- ......
t:: ....
0
....
0..
l_/
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.
1- ~v
I-'"" ....
~----~
1- 10.000 ~ gE<il ~
1- ~ ...... ~ ~ec:.:::
9,000 1Y > (1) 0 (1)
r,l 1/ I-1- :..,.... ......
8.000 o ] eO..N~-<
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·;:::.a
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c-"
o ce ....,
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(1)
.... ..c::: ::;
7.000
u ::; 0..
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1/ 6,000 <
~
..c:::5
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.... <I)
e
/, rl/
o..c::: ....
5,000 ~ \J -~ 15
E = 29.0 X 10 6
1//,rf ~ ~ ~.::::
E = 27.0 X 10 6
--...... ...... rt/1. 4,000
c..>
(1)
t::
0
0
.E5·.;::J-g
E = 24:5 X 10 6 I" VII FIG. HA-l 3,500 •• <;.... g
~ 0
(1)
E •22.8x
r-.. ))
106
3,000
E-< ..C:::(!)p.
.... .... (1)
ell ....
O
~ ~~
0..
El=tl'ii106
I I IT z ·;M::: .s.._. ;:j
FACTOR A
.,...-..,.,...,...=<.__,..-~--~~,~JII:'?:'l'Ci.t~.NN!<~U:"':;;""··R~=-'i!.--- ....t<-... ~-->· •• ):p;,, - · ,...... ---N~-,"·~- , .. R.rit.i- .• *'" 3 j •.• ) ;e•;"" !l!1-.!t'!!~9.~5t-.,.. ....__, j _,. ___ ,_____ .X .. ( ,.,_Q! §..£ -o-AJ.~.-4£ bJI:!
.... .....
..c:::ce..c:::
(1)
:?0.000 .se'O
18.000 ~ ~ s::
UP to ,.!.00 F -ell 0
' 16,000 ~ l'(j ·.;:::
I I " <!)
c..>
400 F,-
~~
,_... ' ''F
1-700
14.000
"" = ......
......
•.. ·-
<I)
0
~I-' 12.000
~-
0 (1) 0..
--
'
/
v ~
,....1- ~ ~ 1-- i-1-
9
l1Fij 10,000 ~ ~B<d~
"Ce~Q;.:::
' / I-' ~ 1,200 F 1Y > (1) 0 (1)
9,000
/ -""'
~---" _... ~
8.000 o""" eO..N~-<
·c .a
]-<Uoce
Ill ""' ~-'V j,...; .....
7,000
c.....
c-"
u
Q) ... ..c::: ....
i)<!><l.)&,
'/.I, 7 6,000
<
~
..c::: 5 ..c:::
~ <;.... ..... (1)
e
f/V 0 ..c::: .....
ell ..... <I)
l'(lJ 5.000 (1) "0 ·- ..c:::
ell;:::~-
E = 28.0 • 106 ,....
IT/. '(f. ""' ce<!> '+-<
c..> t:: 0
E = 25.9 x 1os..J. i- f1'1W 4,000 t::
~-B·.c]
(1) 0
10~~ r"f Vfj
E
E
= 23.8
22.4 X lOS~
X
3 4 5 6 789 2.500
2 3456789 2 3 456789 2 3 456789 2
.00001 .0001 .001 .01 .1
FACTOR A
up to 100 F
!---I - 14.000 C/) ~ s::
f.- ::::::
C/)
ttl ttl--
0
12.000 ~ ... t)
-
...,. <U <U
~ I ""-s:: ....
~ !--" 10.000 •.-~·- .....
0
..... 4001 0 <U 0..
/ ..... .
v t-_ 600 F
9.000
~ <U;j~<U
s::
-
;j ...... ttl
8,00(
CdfS!E~
lr ...,.I-- ........- v BOO F -~ ;> <U 0 <U
...... _......
1.00C 0.. N ,_,
~
.... 6.00(
0
E-!
]...... 8<U ·;:::0 .sttl
........ u <U - ..<::: ~
v 1-o.u<UO..
~
I .-~"'"
] -5 ..<::: 8
/, ......
5.000
1---'
--
,. """ ..... <U
0 ..<::: ....
'/ ,.I--- v ;>-
"' .... <U
4,000 <U ""0 ·- ..<:::
::::17 fl.rf-"
"'
ttl.u s:: ~ """-
E • 28.0 X 106 - !--' 3.500 0 <U s:: 0
E • 25.9 x 106 --.!,..{ s:: ..<::: .9 ""0
"""'"i~......,~
E - 24.5 x 10 -
6 FIG.HA-3 3,000
•• """' ~ <U
~ 0 " ' .....
E : 23.1 X 106 - E-< ...... ..... <U
2.500
o-EhES:
2 3 4
I lII
56789
~
2 3 456789 2 3 456789
ll I 2 3 4 56789
2,000
z ·;::: .5 ::I
FACTOR A
20,000
I I I II ]
up to 100 F
18.000 ..... § ]
.....
-~---
-v i-
,....-
~,...
- J
rI I
.-IoJ
16,000
14.000
12.000 ~
~
0
..... """'
0
.
~
v
1.---
1.---
:,.... 7
....... ~
~..--~
~
I--
1-- ~ '-
- 41 ,OOI F -
1
TF-1
I I I
10,000
9,000 ~
-<
"""0
g:;-:::d)
<U
- .w c
v 8,000 \\o<
n
7 1...-- ~"""
:.....-
[./
~~
c...- - ~~--- ~800 F
7,000
6,000
~
0
E-!
~
;>~§~
.uO..N!-o
..<:!
":j;l.u5"d
~ .w --
8 ·- ::I
[.... ...... 7 U d)-..<::: ....
.... <U <U
_.,...
__. I 5,000 < ]-s<DO..
II ~ I lilot ~""" -s ~
w 1-- 4,000
"' 0
<U ""0 .-;: ]
~ s::
..<::: ......
~ ......
E = 28.0 x 1os-
~~ 3,500 0
s::
d)
d) §
"""
E • 26.4 x 106-
;-... -tt'& 0
- -s · - ""0
FIG.HA-4 3,000
•• """ t) ~
b~
E 24.5 • 1Q6- ~ 0 <U
E 23.1 X lOS ......
2,500 E-<-~~
O"Eh2S:
I II ~ I 2.000
z ·;::: .5 ::I
2 3 456789 2 3 4 56789 2 3456789 2 J 4 56789
.00001 .0001 .001 .01 .1
FACTOR A
CONSTRUCTION
It is preferable to use plates in constructing a composite-section stiffener ring,
rather than using standard structural shapes. The reason for this lies not only in
the difficulties of rolling heavy structural shapes, but also because of the neces-
sity to adjust the ring to the curvature of the shell. For large diameter vessels the
maximum permissible out of roundness can result in a 1 - 2 inch gap between
the shell and the ring. This can be eliminated if the vertical member of the ring is
cut out of the plate in sections. The sections can be flame cut, instead of rolled
and then butt-welded together in place.
DRAIN AND VENT
Stiffener rings placed in the inside of horizontal shells have a hole or gap at the
bottom for drainage and at the top for vent. Practically one half of a 3 inch
diameter hole at the bottom and 1!h inch diameter hole at the top is satisfactory
and does not affect the stress conditions. Figure A.
For the maximum arc of shell left unsupported because of gap in stiffening
ring, see Code Figure UG.29.2.
WELDING
According to the ASME Code (UG 30): Stiffener rings may be attached to the
shell by continuous or intermittent welding. The total length of intermittent
welding on each side of the stiffener ring shall be:
I. for rings on the outside, not less than one half the outside circumference
of the vessel;
2. for rings on the inside of the vessel, not less than one third of the circum-
ference of the vessel.
Where corrosion allowance is to be provided, the stiffening ring shall be attached
to the shell with continuous fillet or seal weld.ASME. Code (UG.30.)
Max. Spacing
12 t for internal ring
8 t fO< <xt<m.U ring l 20 30 40 SO 60 70 80 90 I 00 II 0 120 I 30 140 l SO 160 I 70
1:
(Specified yield strength 30,000 to 38,000 psi, inclusive)
To find ~e required l!ead thickness: 1. Determine R, 2. Enter the chart at the value
of R, 3. Move vertic<illy to temperature line, 4. Move horizontally and read t.
t = Required head thickness, in.
Figure A Figure B R = For hemispherical heads, the inside radius, in.
EXAMPLE: RINGS OUTSIDE W' x 3" lg. fillet weld on 6" ctrs. For 2:1 ellipsoidal heads 0.9xD 0
RINGS INSIDE "" x 2" lg. fillet weld on 6" ctrs. For flangeq and dished heads, the inside crown radius, in. Rmax=Do
D0 = Outside diameter of the head, in.
The fillet weld leg-size shall be not less than the smallest of the following: 1/4 in,
~ ""' ...... • "'· !_!-A.
50 51
CHARTS FOR DETERMINING 'IHE WALL THICKNESS FOR CHARTS FOR DETERMINING 'IHE WALL 'IHICKNESS FOR
VESSELS SUBJECTED TO FULL VACUUM VESSELS SUBJECTED TO FULL VACUUM
..... .... ,10 .1s .ao .z . .ao .35 .<40 •.d!S .so .ss .eo .es . TO .TS .eo .as .liiO ~1ill!5
. ,..,
- '\._",
~
tl± - r--: ...,..,
r--:
-
--
300°f
- ~"'""~"' -
. ..,;, . ..,;,
v - 1-500 Of
1'--.~ ~ _.....!-!-700 Of .,.., i 375.
~ ~v ~soo°F ..., l
~ ~K .,--- 900°f
v .... ..... ..... ....
..,...
~~~
N !::' ~
~
300.
275.
....
0
0 0
Q """
215.
""" ~
~ ~:::,._
.... ...."'""
:100.
,,.., "'-;::
~~ """
,.,., 115,
~~ ,,., ISO.
1... ....
• . 5 • 1 '""•
•
'· • • . 5 • ' • 0 10.
100. tOO.
... 1.1.
0
:X:
NOTATION
Required shell thickness, in.
50.
!- Outside diameter of shell, in.
z
..... w
...)
" Length of the vessel or vessel section, taken as the largest of the following:
l. Distance between the tangent lines of the heads plus one third of the depth of
... II
....1
the hea_ds- if stiffening rings are not used, in.
·2. The greatest distance between any two akjacent stiffening rings, in .
2<).
3. The distance from the center of the first stiffening ring to the head tangent
line plu's one third of the head depth, in.
10.
The charts are from!
"' Logan, P. J., "Based on New ASME Code Addenda •.. Chart Finds Vessel Thickness,"
CYLINDRICAL SHELL HYDRO~ARBON PROC_,ESSING, 55 No. 5, May 1976 p. 217.
(See facing page for explanation) Logan, P. J., "A Simplified Approach to •.. Pressure Vessel Head Design," HYDROCAR·
BON PROCESSING, 55 No. 11, November 1976 p. 265.
Copyrighted Gulf Publishing Co. Houston. Used with permission.
52 53
LI Standard.)
(D x H)
I
Projected area oftower, sq. ft.
height of tower considered, ft.
200
300
500
1.4
1.6
1.9
1.9
2.0
2.3
2.1
2.2
2.4
The area of caged ladder may be approximated as 1 sq. ft. per lineal ft. Projected-area
of platform 8 sq. ft.
outside diameter of tower, ft.
'---Shape factor= 0.8 for cylindrical tower (Table 6-7) Users of vessels usually specify wind pressure for manufacturers without reference
'----Gust response factor (Gh & GJ* (Para. 6.6) to the height zones or map areas. For example: 30 lb. per sq. ft. This specified pres-
sure shall be considered to be uniform on the whole vessel.
When the tower is located:
in urban, suburban areas, Exposure B 0.8; The total pressure on a tower is the product of the unit pressure and the projected
in open terrain with scattered obstruction, Exposure C 0.85; area of the tower. With good arrangement of the equipment, the exposed area of the
lll
in flat, unobstructed areas, Exposure D 0.85.
wind can be reduced considerably. For example, by locating the ladder 90 degrees
'-----Velocity pressure at height z above ground, lb./sq. in. from the vapor line.
0.00256 KzKzc V2 I, lb./sq. ft. (Table 6-1) •'
I ; EXAMPLE:
-Design Wind Force, lb. lmportonce fuctor 1.0 fi" 'tructur"' that Determine the wind load, F
on projected area of present low hazard to human life in event
tower. (Para. 6.2) offailure (Para. 6.2). DESIGN DATA:
the wind speed, V = J<()Om.p.h
Wind speed, mph. (Map 6-1)
Topographic factor= 1.0 when wind speed-up
diameter of tower, D = 6ft.
over hills and escarpment is not present. height of tower, H = 80ft.
_ (Para. 6.5.5) the tower located in flat,
,,
unobstructed area, exposure: D
Velocity Pressure
Exposure Coefficient* The wind load, F=qz xG x C1 AJ
Exposures B, C & D (Table 6-3) · q from table = 26 psf
* See tables below for values of q and for combined values G froq1 table = 1.8
of Gh, Gz, and Kz in Exposures B, C, and D. Shape factor = 0.8
VEWCITYPRESSURE,q Area,A1 ';'DH=6 x 80 480 sq. ft.
Basic wind speed, mph, V I 10 I so I ~ 1100 1110 1120 1130 1 F = 26 X 1.8 X 0.8 X 480 17,97llbs.
Velocity Pressure psf0.00256 V2, q 1 13 1 11 1 21 1 26 1 31 1 37 1 44 1
54 55
90
100
110
120
130
.I
~
Alaska Note:
For coastal areas and islands, Special Wind Region
use nearest contour.
• Population Center
Location V, mph
Hawaii 105
Puerto Rico 125
Guam 170
Virgin Islands 125
American Samoa 125
Notes:
1. Values are 3-second gust speeds in miles per hour itt 33 ft.
above ground for Expo'Sure C category and are associated with
an annual probability of 0.02.
2. Linear interpolation between wind speed contours is permit-
ted.
3. Islands and coastal areas shall use wind speed contour of
coastal area. •
ANSIIAASCE STANDARD 7-95 4. Mountainous terrain, gorges, ocean promotories, and special
Courtesy of American Society of Civil Engineers wind regions shall be examined for unusual wind conditions.
56 57
::s
when the horizontal cross sec- .... "'
:I!~
~
al
-.=.
tion is hexagonal or octagonal .,_ :.::to•
20 30 35 and with 0.60 when the horizon- z :::lo: .Q
tal cross section is circular or el- C(l)::•
25 40 45 50
liptical.
I-LIJ ....
sa::~!
z
(l)~d
0)
25 30 40 45 50 55 60 ..s
1.1.1(1); .....
30 40 45 55 60 70 75 a::w o:::: >.
wa:: I ..... .0
tSQ. "•Vo
.....
EXAMPLE: 0"''"'
~z;i ..
Find the wind pressure Pw from map. oj!l'l'"
" ..J oi"'
•
_, ~=;
The vessel is intended to operate in Oklahoma, which is in the wind pressure map area ct ;:~
marked 30. In this map area the wind pressures for various height zones are: :::u
In the height zone less than 30 ft. 25 lb. per sq. ft.
In the height zone from 30-49 ft. 30 lb. per sq. ft. ...0)
For a cylindrical tower these values shall be multiplied by shape factor 0.6, then the ..so:l
0)
wind pressure in different zones will be 15 and 18 lb. per sq. ft. respectively
:=::
If many pieces of equipment are attachfd to the tower it is advisable to increase the ..."'
-
0)
o:l
shape factor (according to Brownell) up to 0.85 for a cylindrical vessel. •' !:1:)
Users of vessels usually specifY the wind pressure for manufacturers without refer- "'- 0)
ence to height zones or map areas. For example: 30 lb. per sq. ft. This specified pressure
shall be considered to be uniform on the whole vessel.
·=
;_:)
II>
Relation between wind pressure and wind velocity, when the horizontal cross section ..s
is circular, is given by the formula: . .....
0
EXAMPLE: ..
~tit~~
"'
0
Q' cz ...
II>
II>
• 0
z -
> ::> ..s
Wind of 100 mph velocity exerts a pressure:
Pw 0.0025 x Vw 2 = 25 lbs. per sq. ft. pressure on the projected area of a cylindrical
: i : a
c.
0
~
= 0
"'..~
lit •
c. 0 • ...
vessel at a height of 30 feet above ground. 1- 0
z ! ..J ""
j <C
c :z: 0 <C .0
0 u u •
The total wind pressure on a tower is the product of the unit pressure and the projected •• •••••• 0.
o:l
area of the tower. With a good arrangement of equipment the exposed area of the wind ;' • • •• s
can be reduced considerably. For example, by locating the ladder 90 degrees from the II>
vapor line. ~
58 59
iI
DESIGN OF TALL TOWERS ' DESIGN OF TALL TOWERS
WIND LOAD
(Continuation) WEIGHT OF THE VESSEL
FORMULAS
SHEAR MOMENT . REQUIRED The weight of the vessel results compressive stress only when eccentricity does not
STRESS THICKNESS exist and the resultant force coincides with the axis of the vessel. Usually the
compression due to the weight is insignificant and is not controlling.
The weight shall be calculated for the various conditions of the tower as follows:
+-~j""'"'m
Given: D1 = 3ft. 6 in. H = 100ft. 0 in. hT = 4ft. 0 in. s= w where S = unit stress, psi
Pw = 30 psf ct W = weight of vessel above the section under consideration, lb.
Determine the wind moment c = circumference of shell or skirt on the mean diameter, in.
~ r-- h1 = H/2 = 50 ft. 0 in.
"0 t = thickness of the shell or skirt, in.
"0
PwxD 1 xH= Vxh,= M
._"'
..l
Vessel 30 x 3.5 x 100 = 10,500 x 50 = 525,000
0
Q, Ladder 30 x 98 lin. ft. = 2,940 = 49 = 144,060
The weight cif different vessel elements are given in tables beginning on page 374
~ Platfonn 30 X 8 lin. ft. = 240 x 96 = 23,040
Total V = 13,680 M = 692,100
Moment at the bottom tangent line ft. lb
MT = M- hT(V- 0.5 PwDI hT) =
692,100 - 4 (13,680 - 0.5 X 30 X 3.5 X 4) = 638,220
ft. lb.
SEE EXAMPLES FOR COMBINED LOADS ON PAGE: 69
61
NOTATION
The base shear is the total horizontal seismic shear at
D = Outside diameter of vessel, ft. the base of a tower. The triangular loading pattern and
H= Length of vessel including skirt, ft. the shape of the tower shear diagram due to that load-
g = 32.2 ft. per sec. squared, acceleration ing are shown in Fig. (a) and (b). A portion of F1 of total
t = Thickness of skirt at the base, in. horizontal seismic force Vis assumed to be applied at
v = Total shear, lb. CW, see page 61 (a) Seismic Loading Diagram the top of the tower. The remainder of the base shear is
:i,, distributed throughout the length of the tower, includ-
W= Weight of tower, lb.
ing the top.
w = Weight of tower per foot of height, lb.
Overturning Moment
EXAMPLE
The overturning moment at any level is the algebraic
Given: Determine the actual and maximum allowable sum of the moments of all the forces above that level.
period of vibration
D = 3.125 ft. 0 in.
H = 100ft. 0 in. NOTATION
g = 32.2 ft/sec 2 . I coeffiICient
. ~ 2.3 5S
C = Numenca
t = 0.75 in.
v = 1440 lb.
r=o.oooo265eoo~
3.125
"36ox3.125 = 1.05 sec.
0.75 (need not exceed 2.75)
W= 36,000lb.
.y 36000
w
C =Numerical coefficient= 0.035
in operating condition X I 00
Ta= O.so X . =7.05 sec.
w = 360
' 1440 32 2 D =Outside diameter of vessel, ft.
E =Efficiency of welded joints
The actual vibration does not exceed the allow- . F1= Total horizontal seismic force at top of the vessel,
able vibration. (b) Seismic Shear Diagram lb. determined from the following formula:
F,=0.07 TV (F,need not exceed 0.25 V)
Base Shear
=O,forT 50.7
H =Length of vessel including skirt, ft.
Reference: Freese, C. E.: Vibration of Vertical Pressure Vessel ASME Paper 195 9.
62 63
DESIGN OFTALL TOWERS DESIGN OF TALL TOWERS
Given:
NOTATION
·- Seismic zone: 2B Z=0.2
' I = Occupancy importance coefficient (use 1.0 for
vessels) D= 37.5 in.= 3.125 ft. X= 96ft,. 0 in.
M =Maximum moment (at the base), ft-lb. H= 100ft., 0 in. W=35,400 lb.
Mx =Moment at distance X, ft-lb. Determine: The overturning moment due to earthquake at the base and at a
R = Mean radius of vessel, in. distance X from top tangent line.
Rw =Numerical coefficient (use 2.9 for vessels) First, fundamental period of vibration shall be calculated.
s = Site coefficient for soil characteristics T=C1 xH%=0.035 x IOO%= 1.1 sec.
D A soil profile with either: and
a) A rock-like material characterized by a shear-wave I= 1, S= 1.5, Rw=2.9,
,....... velocity greater than 2,500 feet per second or by
other suitable means of classification. S = 1.0
C= 1.25S = 1.25 X 1.5 = 1.76<2.75
b)Stiff or dense soil condition where the depth is r2f3 I.J2f3
less than 200 ft. S = 1. A soil profile with dense or
stiff soil conditions, where the soil depth exceeds
200 feet. S = 1.2. V= ZIC x W= 0.2x 1 x 1.76 x35,400=4,296lb.
-r- Rw 2.9
X A soil profile of 40 feet or more in depth and con-
H taining more than 20 feet of soft to medium stiff
2
clay, but not more than 40 feet of soft clay. S = F 1 = 0.07 TV = 0.07 x 1.1 x 4,296 = 330 lb.
hH 1.5.
A soil profile containing more than 40 feet of soft M= [F1 H + (V- F 1) (2HI3)] =
clay. S = 2.0.
[330 X 100+(4,296 -330)(2 X 100/3)) =294,756ft. -Jb.
--~ st = Al.lowable tensile stress of vessel plate material,
~
psi.
X> H thus
T = Fundam~tal period of vibration, seconds 3
=Ct X H 4 Mx = [Ft X+ ( V-Ft) (X- H/3)] =
t = Required corroded vessel thickness, in. [330 X 96 + (4,296- 330)(1 00-33)] = 281,138 ft. -Jb.
12M or 12Mx
nR2S1E nR2 S1 E
v = Total seismic shear at base, lb.
W = Total weight of tower, lb.
X =Distance from top tangent line to the level un-
der consideration, ft.
z = Seismic zone factor, ,t
0.075 for zone 1 I
0.15 for zone 2A
0.2 for zone 2B i,~~:..
0.3 for zone 3 !i
l
.
0.4 for zone 4
(see map on the following pages for zoning). '
'
~
0\
.!>-
-
Cll
~
-
en
s:::
C":l
N
~~
s:::
?;
0
~
~
~
e
~
~
0
Cll
""'
~
Cll
For areas outside of the United States, see Appendix Chapter 23 of UBC :1991
z
Q
~
DESIGN
66 67
A tower under axial compression may fail in two ways because of instability:
1. By buckling of the whole vessel (Euler buckling)
2. By local buckling
Towers and their internal equipment are usually symmetrical around the vertical In thin-walled vessels (when the thickness of the shell is less than one-tenth of
axis and thus the weight of the vessel sets up compressive stress only. Equipment the inside· radius) local buckling may occur at a unit load less than that required
attached to the vessel on the outside can cause unsymmetrical distribution of the to cause failure of the whole vessel. The out of roundness of the shell is a very
loading due to the weight and result in bending stress. This unsymmetrical arrange- significant factor in the resulting instability. The formulas for investigation of
ment of small equipment, pipes and openings may be neglected, but the bending elastic stability are given in this Handbook, developed by Wilson and Newmark.
stresses exerted by heavy equipment are additional to the bending stresses resulting Elements of the vessel which are primarily used for other purposes (tray
from wind or seismic load. supports, downcomer bars) may be considered also as stiffeners against buckling
if closely spaced. Longitudinal stiffeners increase the rigidity of the tower more
effectively than circumferential stiffeners. If the rings are not continuous around
the shell, its stiffening effect shall be calculated with the restrictions outlined in
the Code UG-29 (c).
FORMULAS
FORMULAS
~.
REQUIRED
MOMENT STRESS THICKNESS ALLOWABLE STRESS (S)
Without Stiffener With Stiffener
M= We S- 12We
- 7! R 2t
12We
t = R 2 1rSE
S= l,SOO,OOO~(<}yield point) S~ l,SO~,OOO v"'t;t; (< j yield p,)
. I ~--------~----------~------------
NOTATIONS:
l i
e =
NOTATION
&x:entricity, the distance from the tower axis to center of
Ax
Ay
= Cross sectional area of one logitudinal stiffener, sq. in.
= Cross sectional area of one circumferential stiffener, sq. in.
a. = Distance between logitudinal stiffeners, in.
£, = Distance between circumferential stiffeners, in.
I w E
M
eccentric load, ft.
Efficiency of welded joints.
= Moment of eccentric load, ft. lb.
K
S
Mean radius of the vessel, in.
= Allowable compressive stress, psi
= Thickness of shell, in.
R = Mean radius of vessel, in. t + ~ The equivalent thickness of the shell when longitudinally
s Stress value of material, or actual bending stress, psi 1
x dx stiffened, in.
t = Thickness of vessel, excluding corrosion allowance, in.
w &x:entric load, lb.
1
~ The equivalent thickness of the shell when circumferentially
Y = t + dy stiffened, in.
COMBINATION OF STRESSES
Towers should be designed to deflect no more than 6 inches per I 00 feet of height.
The.'deflection• due to the wind load may be calculated by using the formula for The stresses induced by the previously described loadings shall be investigated in
uniformly loaded cantilever beam. combination to establish the governing stresses.
Combination of wind load (or earthquake load), internal pressure and weight of
FORMULA
the vessel:
Stress Condition
At windward side At leeward side
+ Stress due to wind Stress due to wind
+ Stress due to int. press.. + Stress due to int. press.
NafATIONS - Stress due to weight - Stress due to weight
= Maximum deflection (at the top), in. Combination of wind load (or earthquake load), external pressure and weight of
= Width of the tower with insulation, etc. ft. ,(
the vessel:
H = Modulus of elasticity, psi
= Length of vessel, included skirt, ft. Stress Condition
= R3-rr t, moment of inertia for thin cylindrical shell
At leeward side
At windward side
(when R> lOt)
= Mean radius of the tower, in. · + Stress due to wind - Stress due to wind
= Thickness of skirt, in. Stress due to ext. press. Stress due to ext. press.
= Wind pressure, psf Stress due to weight - Stress due to weight
The positive signs denote tension and the negative signs denote compression. The
summation of the stresses indicate whether tension or compression is governing.
EXAMPLE It is ·assumed that wind and earthquake loads do not occur simultaneously, thus
Given: the tower should be designed for either wind or earthquake load whichever is
Determine the maximum deflection: !J.M
greater.
D 1 = 2ft., 6 in.
E = 30,000,000 Bending stress caused by excentricity shall be summarized with the stresses
H =48ft.,Oin. resulting from wind or earthquake load.
I = R3 -rr 0.3125
P.., =30psf The stresses shall be calculated at the following locations:
30 X 2.5 X 48 (12 X 48)3
R = 12 in. 69 1. At the bottom of the tower
!J.M = 8 X 30,000,000 X J23 X 3.14 X 0.3125 = 1. in.
t = 0.3125 in. 2. At the joint of the skirt to the head
3. At the bottom head to the shell joint
4. At changes of diameter or thickness of the vessel
The maximum allowable deflection 6 inches per I 00 ft. of height:
The stresses furthermore shall be examined in the following conditions:
for 48'-0" = ---wn-
48 X 6
= 2.88 in. 1. During erection or dismantling
· 2. During test. ·
Since the actual deflection does not exceed this limit, the designed thickness of the skirt is 3. During operation
satisfac..tory. Under these· different conditions, the weight of the vessel and consequently, the
stress conditions are also different. Besides, during erection or dismantling the
vessel is not under internal or external pressure .
1~
For analyzing the strength of tall towers under various loadings by this
A method for calculating deflection, when the thickness of the tower is not con- Handbook, the maximum stress theory has been applied.
stant, given by S. S. Tang: "Short Cut Method for Calculating Tower Deflection". . '.
I•
Hydrocarbon Processing November 1968.
!
70 71
1
~
for internal pressure is available to resist the bending force of the
wind. From Table A, using factor m can be found the distance X
down from the top tangent line within which the thickness calcu-
""
II
::t .c.,. pw = 30 psf wind pressure
R = 12 in. inside radius of vessel
s = 15700psi stress value of SA 285 C
..~l""
II
lated for internal pressure satisfactory also to resist the wind
pressure. ..::
...
..::
material at 200°F temperature
v = Total shear lb .
X =H X m
No allowance for corrosion.
tP = The required thickness for internal pressure
(Hoop Tension) in.
t,. = The required thickness for wind pressure at the bottom head Minimum required thickness for internal pressure considering the strength of the long seams:
joint to shell, in.
I,, _fli 250 X 12 _ 3,000 = O .
EXAMPLE: ~ = 0.233 in., tw = 0.644 in. tjt = 0.644/0.233 = 2.7 228
t = SE- 0.6P = 15700x 0.85 0.6 X 250- 13,195 . m.
tf = 100 ft. p
From Thble m = 0.43 and X = mH = 0.43 x 100 = 43 ft. Minimum required thickness for internal pressure considering the strength of the girth seams:
- o.o
Figure B shows the moment diagram of a tower under wind
_fli _ 250 X 12 = 3,000 =O
112
,
0.1- pressure. The diagram can also be used to select the appropriate
t = 2SE + 0.4P - 2 X 15,700 X 0.85 + 0.4 X 250 26,790 ' m.
plate thickness at various heights. I
Required thickness for longitudinal bending due to wind pressure. Moment at the base (M):
0.21t P.., X D1 X H = V X h1 = M
EXAMPLE: li 30 X 2.5 X 48 = 3,600 X 24 = 86,400 ft. lb.
::t
0.3lJJ- At the height ofO. 71 H the required thickness is 0.5
times the thickness required at the bottom. Moment at the bottom seam (Mr)
t:i
1.) 0.4
\ If the required thickness is: Mr = M - hr (V - 0.5 P.., D 1 hr) = 86,400 - 4 (3,600 - 0.5 X 30 X 2.5 X 4)
~ \ for internal pressure, tP = 0.250 in. = 86,400 - 13,800 = 72,600 ft. lb. = 72,600 x 12 = 871,200 in. lb.
~ o.s !\. I for wind load, t.., 0.625 in.
tf\
at the bottom required Required thickness:
"'
0
0.6 f/2 + fw =0.750 in.
- .-.!.1:.x._ -
""
::t
c.:J at height 0.71 H; I t - R2 'IT SE - 122 X
871 ,200
3.14 X 15,700 X 0.85
871,200 0 145 .
6,037,135= · m.
w 0.7
::t
0.5 X 0.750 = 0.375 in.
"'
thickness for internal The required thickness calculated with the strength of the bottom girth seam:
0.8 pressure t/2 = 0.125 in. For wind pressure 0.145 in-
required thickness at 0.71 H = 0.500 in. Forint. pressure 0.112 in.
This is greater than the thickness calculated with
0.9 r--
IT Tai'AL 0.254 the strength of the longitudinal seam therefore, this
- 1.0
o.t 0.2 o.3 o.4 o.s
Ratio of plate thickness required at the bottom
(t 12 + t) to thickness required at the consid-
eted height.
0.6 0.1 o.a 0.9 1.0 Fig. B
,
·'
..
:·
,.
.••.
minimum thickness 0.257 in. shall be used .
For simple vessels where the moment due to wind is small, th~ abov~ c~lculat!on is satisfactory.
Vessels which are subject to larger loadings may need closer mvest1gat1on w1th respect also to
.•
economical viewpoints. See pages 76-84 for skirt, base and anchor bolt design.
72 73
9 3'-6"
0 .
0....
v..., ~
and for internal pressure (hoop tension) only 0.25 plate
is satisfactory. For economical reasons it is advisable to
~ ~Platform
c f-- use different plate thicknesses at various heights of the
DESIGN DATA tower.
D 3 ft. 0 in. inside diameter f--
r-~/ D, 3 ft. 6 in. width of vessel with insulation, allowance for
piping, etc.
-
1-- f--
The thickness required for hoop tension (0.25 in.) serves
to resist also the wind load to a certain distance down
~
. ...0.
r- f-- from the top,
0
.
"C E 0. 85 efficiency of welded seams
hr = 4 ft. 0 in. distance from the base to the bottom head to shell
~
"'c::i - Find this distance (X) from table A, Page 70
~
...:1
.... =
9 ""
0 H
joint.
I00 ft. 0 in. length of tower
- tw/tp = 0.564/0.204 = 2.7 then X= 0.43 x H 43 ft .
From diagram B, Page 70 can be found the required
§ ,.: r--
9 P
Pw
= ISO psi internal pressure
30 psf wind pressure 0 thickness and length of the intermediate shell sections.
~ R = 18 in. inside radius of vessel ...... "'.,..c::i
f-- Using 8 ft. wide plates, the vessel shall be constructed
II
::r:
•9 S = 15700psi stress value of SA-28SC material at 200"F
f-- from:
0
.,.., (5) 0.25 thick 8 ft. wide courses
II V
temperature
Total shear, lb.
0
;,.
..... (4) 0.50 thick 8 ft. wide courses
40ft.
32ft.
Head: 2: I seamless elliptical ......... (3) 0. 75 thick 8 ft. wide courses 24 ft.
~r~
~
~~
s,--= w 34,000 v
Co
0
Shell 30 X 3.5 X 40 = 4,200 X 20 = 84,000
in operating condition 392 psi II
~
Cmt 115.5 X 0.75 X "' "' Platform 30 X 8 lin. ft. = 240 X 36 = 8,640
COMBINATION OF STRESSES Ladder 30 X 38lin. ft.= 1,140 X 19 = 21,660
WINDWARD SIDE I LEEWARD SIDE Total Moment M"
IN EMPTY (ERECTION) CONDITION 12 Mr = 12 X 114,300
[. S= R2 1T t = 5,316 psi
18.1252 X 3.14 X 0.25
Stress due to wind + 9,640 Stress due to wind - 9,640
Stress due to weight - 358 Stress due to weight -
---358 Stress due to internal pressure
(As calculated previously) 1,837 psi
+ 9,282 psi - 9,998 psi t:::t: Total 7,153 psi
(No int. pressure during erection)
IN OPERATING CONDITION
Stress due to int. press. + 1,837 Stress due to wind 9,640
Stress due to wind + 9,640 Stress due to weight - 392
The 0.25 in. thick plate for shell at 40 ft. distance from top of the tower is
+ 11,477 ~ 10,032
Stress due to weight - 392 Stress due to int. press. + 1,837 satisfactory. No further calculation is required on the same reason mentioned above.
+ 11,085 psi - 8,195 psi i
i
The tensile stress 11,085 psi in operating condition on the windward side governs.
The allowable stress for the plate material with 0.85 joint efficiency is 13,345 psi.
Thus the selected 0.75 in. thick plate at the bot.tom of the vessel is satisfactory.
Stress in the shell at 72ft. down from the top of tower. Plate thickness 0.50 in.
J""""'. Stress due to wind.
q
d-Lr-- ...
Pw XD I XX=VX-=M
X
2 X
N
r- .
9 ~
Shell
Platform
30 X 3.5 X 72 = 7,560 X 36 = 272,160
30 x 8 lin.-ft. = 240 X 68 16,320
II g Oo
'-0
Ladder 30 X 70 lin.-ft, = 2,100 X 35 73,500
>< Total Moment M,. = 36T,980 ft.-lb.
12M 12 X 361,980
s = R2 1T t 18.252 X 3.14 X 0.50
8,303 psi
Stress due to internal pressure
(As calculated previously) 1,837
::!=1: Total 10,140 psi
The calculation of stresses at the bottom head has shown that the stresses on the
windward side in operating condition govern and the effect of the weight is insig-
nificant. Therefore without further calculation it can be seen that the tensile stress
10,140 psi does not exceed the allowable stress 13,345 psi. Thus the selected 0.50
in. thick plate is satisfactory.
76 77
A skirt is the most frequently used and the most satisfactory support for vertical Vertical vessels, stacks and towers must be fastened to the concrete foundation,
vessels. It is attached by continuous welding to the head and usually the required skid or other structural frame by means of anchor bolts and the base (bearing)
size ofthis welding determines the thickness of the skirt. ring.
Figures A and B show the most common type of skirt to head attachment. In the The number of anchor bolts. The anchor bolts must be in multiple of four and
calculation ofthe required weld size, the values ofjoint efficiency given by the Code for tall towers it is preferred to use minimum eight bolts.
(UW12) may be used.
i Spacing of anchor bolts. The strength of too closely' spaced anchor bolts is not
i:
I j fully developed in concrete foundation. It is advisable to set the anchor bolts not
FORMULA I closer than about 10 inches. To hold this minimum spacing, in the case of small
12MT W diameter vessel the enlarging of the bolt circle may be necessary by using conical
t= R2 1tSE + D skirt or wider base ring with gussets.
Diameter of anchor bolts. Computing the required size of bolts the area within
NOTATIONS the root of the threads only can be taken into consideration. The root areas of
D = Outside diameter of skirt, in. · bolts are shown below in Table A. For corrosion allowance orte eighth of an inch
E = Efficiency of skirt to head joint. should be added to the calculated diameter of anchor bolts.
(0.6 for butt weld, Fig. A, 0.45 for lap weld, Fig. B) For anchor bolts and base design on the following pages are described:
M = Moment at the skirt to head joint, ft. lb.
R T= Outside radius of skirt, in. 1. An approximate method which may be satisfactory in a number of cases.
S = Stress value ofthe head or skirt material whichever 2. A method which offers closer investigation when the loading conditions and
is smaller, psi. other circumstances make it necessary;
t = Required thickness of skirt, in.
W = Weight of the tower above the skirt to the head . TABLE B
joint, in operating condition.
Bolt •
Root Area
sQ. in.
ttl
Dimension in.
12 IJ
NUMBER OF ANCHOR BOLTS
Diameter of
Bolt circle in.
24
42
to
to
36
54
Minimum Maximum
4
8
4
8
60 to 78 12 12
EXAMPLE ~ 0.126 7/8 5/8
84 to 102 12· 16
Given the same vessel considered in Example B.
% 0.202 I 3/4
108 to 126 16 20
% 0.302 1-1/8 13/16
132 to 144 20 24
D 37.5 in. S = 15,700 stress value % 0.419 1-1/4 15/16
E 0.60 for butt joint of SA- 285- C plate 1 0.551 1-3/8 1-1/16
1Ys 0.693 1-1/2 1-1/8
Mr= 638,220 ft. lb. w = 31,000 lb. 17,4 0.890 1-3/4 1-1/4
R 18.75 in. TABLE C
1% 1.054 l-7/8 1-3/8 MAXIMUM ALLOWABLE STRESSES FOR
Determine the required skirt thickness. 1~ 1.294 2 1-1/2 BOLTS USED AS ANCHOR BOLT
1% 1.515 2-1/8 1-5/8 Specification Max. allow.
For wind:
12 MT 12 X 638,220
=0.736in.
1% 1.744 2-1/4 1-3/4 Number
Diameter in.
Stress psi.
t= R2 1tSE + 18.75 2 X3.14X 15,700X0.6 1% .2.049 2-3/8 1·7/8
2 . 2.300 2-1/2 2 SA307 All diameters 15,000
31 000 2;4 3.020 2-3/4 2·1/4 SA 193 B 7 2\.'S and under 19,000
For weight:
3.75X3.14X 15700X0.6
=0.028in. 272 3.715 3-1/16 2·3/8 SA 193 Bl6 2Y, and under 17,000
2% 4.618 3-3/8 2-5/8 SA 193 B 7 Over 2 \IS to 4 incl. 18,000
TOTAL =0.764in. SA 193 Bl6 Over 2 \IS to 4 incl. 15,000
Use 13/16" thick plate for skirt.
3 5.621 3-5/8 2-7/8
REFERENCES: Thermal stresses are discussed in tl1ese works:
Brownell, Lloyd E., and Young, Edwin H., "Process Equipment Design,".Jolm Wiley and Sons, Inc., 1959. Weil, * For bolts with standard threads.
N.A., and J. J. Murphy Design and Analysis of Welded Pressure Vessel Skirt Supports. Asme. Trans. Industrial Engi·
neering for fndustry, Vol. 82, Ser. B., Feb., 1960.
78 79
- .§
~--~-·
-+ e~ .:1~·}·~ anchor bolts. See page 77 Table B Tensile stress in anchor bolts, Sa, psi. s _...!2_
·-l,rC,
·GJ
3. Determine the inside and outside diameter of the
~ I~ e base·ring
~
4. Check the stresses in the anchor bolts and Thickness of a ring which has an B,
area equal to tile area of anchor 1
r\ foundation bolts, ts, in. ' = nd
5. If the deviation between the allowable and
D·kD kD actual stresses are too large, repeat the
Compression load on the concrete,
D calculation F,=F,+ W
Fe. lb.
6. Calculate the base ring thickness
I!
Relationship between tension in steel
r TABLED
Values of Constants TABLE F and compression in concrete. s. = nj;
i ts
as Functions of K Bending moment per unit length of section of
! k Cc Ct j z a plate perpendicular to X and Y axes respeo- ! Base ring thickness without gusset
w
tively. Use greater value, Mx or My. WL/~ Is= /l,j3J;)S
plate, tB, in.
I 0.00
.05
0.000
0.600
3.142
3.008
0.750
.760
0.500
11~b
•
=~
.10 0.852 2.887 .766 .490 Mx My Base ring thickness with gusset
.480
.15 1.049 2.772 .771 .469 plate, tB, in. ls S
r .20
.25
1.218
1.370
2.661
2.551
.776 .459 ~~~-+------------4---------~--
.779 .448 0.000 0.000 - 0.500/cli
.30 1.510 2.442 .781 .438
.35 1.640 2.333 .783 .427 0.333 0.0078j,b 2 - 0.428/c 11 NOTATION
.40 1.765 2.224 .784 .416
.45 1.884 2.113 .785 .404 0.500 0.0293 .fc b 2 - 0.319/c li b = The distance between gusset plates, measured on arc of bolt circle in.
.so 2.000 2.000 .785 .393
B, = Total area required for anchor bolt sq. in.
.55
.60
2.113
2.224
1.884
1.765
.785 .381 0.667 0.0558/c b 2 - 0.227 .fc ll
.784 .369 Cc,C, = Constants, see Table D on the preceding page.
.65
.70
2.333
2.442
1.640 .783 .357 1.000 0.0972 .fc b2 - 0.119 .fc ll d = Diameter of anchor bolt circle, in.
1.510 .781 .344
.75
.80
2.551 1.370 .779 .331 1.500 0.123 .fc b 2 - 0.124/c b 2 D = Diameter of anchor bolt circle, ft.
2.661 1.218 .776 .316
.85
·.90
2.772
2.887
1.049
0.852
.771
.766
.302
.286
2.000 0.131 .fc b 2
- 0.125 .fc b2 fc = Compressive stress in the concrete at the outer edge of the base ring, psi.
.95
1.00
3.008
3.142
0.600
0.000
.760
.750
.270
.250
3,000 0.133 .fc bl - 0.125/c b 2 feb = Compressive stress in the concrete at the bolt circle, psi.
lc bl j = Constant, see Table D on the preceding page.
- 00 0.133 - 0.125/c b 2
.
14 = I - t, in. = width of the base ring, in.
TABLE E M = :"Moment at the base due to wind or earthquake ft. lb.
Properties of Concrete Four Mixtures NOTE: M,..ax = M, or M,, whichever is greater. See Table F on the preceding page.
Ultimate 28 day
Se.e notation on facing page. n = Ratio ofmodulus of elasticity of steel and concrete.Es/Ec. See Table E.
Strength psi 2000 2500 3000 3750 r = Radius of bolt circle, in.
Allowable compr. sa = Allowable tensile stress on anchor bolts, psi.
Strength fc psi 800 1000 1200 1500 s = Maximum allowable stress value of base plate, psi.
Safe bearing load 500 625 750 938
w = Weight of the:; tower at the base, lb.
fb psi z = Constant. See Table D on the preceding page.
Factor n 15 12 10 8
82 83
DESIGN OF ANCHOR BOLT AND BASE RING DESIGN OF ANCHOR BOLT AND BASE RING
EXAMPLE
EXAMPLE (Cont.)
Checking value of kwhich was calculated with assumed values offc:h = 1,000 psi and
DESIGN DATA: DETERMINE: Sa= 18,000.
iJ · "" 5 ft, 0 in. diameter of anchor bolt circle. The size and number of Then the constants from
d "' 60 iri. diameter of anchor bolt circle. anchor bolts; TableD are:
=0.19
n = lO, ratio of modulus of elasticity of steel The width and thickness + 17,960 Cc = 1.184
and concrete (Table E. Page ~0) of base ring. IOX430 c, = 2.683
fc = 1,200 psi allowable compr. strength of j 0.775
concrete (Table E, Page 80) z 0.461
S = 15,000 psi allowable stress value of base M-WzD 692,100-36,000X0.461 X5 =
ring. b
F 157 1921
sa = 18,000 psi allowable tensile stress in'bolts. I JD 0.775 X 5 ' .
W 36,000 lb. weight of the tower.
M 692,100 ft. lb. moment at the base. 157,192 624 .
0.125 X 30 X 2.683 = 15 • pst
SOUJTION:
Assume 8 in. wide base ring and a compressive stress at the bolt circle, feb = 1,000 psi. ,,'!
Then the constants from
TableD are: ~ ~: Fe= F,+ W= 157,192+ 36,000= 193,192lb.
i_t4
k = - - - - " - - - = 0.35
1 + 18,000
cc = 1.640 '
I" -~ Fe 193,192 ,
l + ~ c, = 2.333 i
Jch {!5f n~,) rCc (7.875 + 10 X 0.125) 30 X 1.184 = 596 pst
nfcb lO X 1,000 j "" 0.783
z "" 0.427
Compressive stress in the anchor bolts:.
This is in sufficient agree-
2 X 0.35 X 60 ment with the assumed Sa=nfc"= 10X596 5,960psi
feb = fc kd j; l = 1,200 X 0. 35 X 60 X 8 = 1,008 psi
2 2 value offcb = 1,000 psi Compressive stress in the concrete at the outer edge of the base ring:
Required area of anchor bolts 2kd+ 1 2XO.l9X60+8 .
. 12M - Wzd 12 X 692,100 - 36,000 X 0.427 X 60 SO . fc:=fch X 2kd =596X 2 X 0 . 19 X 6 0 =805psi
23
B, = 2 'II' C Sa jd = 6 '28 2.333 X 18,000 X 0.783 X 60 = . sq. m.
1
Using 12 anchor bolts, the required root area for one bolt Required thickness of base ring /1 6 in.
23.50/12 1.958 in. ,[
_ ~- j3X805 .
From Table A 17/a in. diameter bolt would be satisfactory but adding \Is in. for corrosion,
tJJ -I, VJ fc:/ S- 6 15 OOO 2.406 m.
i' '
use (12) -2 in. diameter anchor bolts. •1:·'
Tensile load on the anchor bolts •;i
To decrease the thickness of the base ring, use gusset plates.
Using (24) gusset plates, the distance between the gussets:
F = M - Wz D = 692,100 - 36,000 X 0.427 X 5 = 157 150 lb.
1
jD 0.783 X 5 '
·~ 6
b= trd=1 85"·!! = --=0 764
. 'b 7.85 .
~'
Tensile stress in the anchor bolts
1
s I Mm~ My;:0.196fc: l/,;;0.196 X 805 X 62= 5680 in. lb.
V6 1~.~~~
'
t = B 23.50
= _..::;;_;,;;,..:;__ 0 125 .
'
--=./.._
'II' d 3.14 x 60 = · m. t8 1.5076 in. Use 1Y:z in., thick base plate.
I,
:IJ
Compressive load on the concrete: 14 = l - t, = 8.0 0.125 7.875 in. 11
!~
F 193,150 . !jl
feb = (l + nt) r C = (7 .875 + lo X 0.125) 30
4 s e
X 1.640 = 430 pSl
·.!~.:
II
84 85
NOTES
The.s:hairs are designed for the maximum load which the bolt can transmit to them.
The anchor boft size and base plate shall be calculated as described on the fore-
going pages.
All contacting edges of the plates shall be welded with continuous fillet weld. The
leg size of the fillet weld shall be one half of the thinner joining plate thickness.
,,
'
l!z"
1'-(J'
DIMENSIONS inches
Anchor A B c D E F G
bolt diam
The above table is taken from Scheiman A.D. Short Cuts to Anchor Bolting and
Base Ring Sizing.Petroleum Refiner, June 1963.
87
86
LOCATION OF SADDLES.
The use of only two saddles is preferred both statically and economically over
the multiple support system, this is true even if the use of stiffener rings is
STRESSES IN LARGE necessary. The location of the saddles is sometimes determined by the location
HORIZONTAL VESSELS of openings, sumps, etc., in the bottom of the vessel. If this is not the case,
SUPPORTED BY SADDLES then the saddles can be placed at the statically optimal point. Thin walled
vessels with a large diameter are best supported near the heads, so as to utilize
the stiffening effect of the heads. Long thick walled vessels are best supported
where the maximal longitudinal bending stress at the saddles is nearly equal to the
The design methods of supports for horizontal vessels are based on L. P. Zick's stress at the midspan. This point varies with the contact angle of the saddles. The
analysis presented in 1951. The ASME published Zick's work (Pressure Vessel distance between the head tangent line and the saddle shall in no case be more than
and Piping Design) as recommended practice. The API Standard 2510 also refers 0.2 times the length of the vessel. (L)
to the analysis of Zick. The British Standard 1515 adopted this method with
slight modification and further refinement. Zick's work has also been used in Contact Angle 9
different studies published in books and various technical periodicals. The minimum contact angle suggested by the ASME Code is 120°, except for
very small vessels. (Code Appendix 0-6). For unstiffened cylinders under exter-
The design method of this Handbook is based on the revised analysis mentioned nal pressure the contact angle is mandatorily limited to 120 o by the ASME Code.
above. (Pressure Vessel and Piping; Design and Analysis, ASME, 1972) (UG-29). .
A horizontal vessel on saddle support acts as a beam with the following deviations: Vessels supported by saddles are subject to:
1. The loading conditions are different for a full or partially filled vessel. 1. Longitudinal bending stress
2. Tangential shear stress
2. The stresses in the vessel vary according to the angle included by the saddles.
3. Circumferential stress
3. The load due to the weight of the vessel is combined with other loads.
LOADINGS:
2. Internal Pressure. Since the longitudinal stress in the vessel is only one half of
the circumferential stress, about one half of the actually used plate thickness
is available to resist the load of the weight.
3. External pressure. If the vessel is not designed for full vacuum because vacuum
occurs incidentally only, a vacuum relief valve should be provided especially
when the vessel outlet is connected to a pump.
4. Wind load. Long vessels with very small t/r values are subject to distortion
'from wind pressure. According to Zick "experience indicates that a vessel
designed to 1 psi. external pressure can successfully resist external loads en-
countered in normal service."
5. Impact Loads. Experience shows, that during shipping, hardly calcul11.ble im-
pact loads can damage the vessels. When designing the width of the saddles
and the weld sizes, this circumstance is to be considered.
88 89
NOTATION:
-~ UJJ·~
L=+JH
Q = Load on one saddle lbs.
R = Radius of shell Positive values .denote tensile stresses and negative values denote compression.
s =Stress pound per sq. inch
t 8 =Wall thickness of shell E =-Modulus of elasticity of shell or stiffener ring materia). pound per square inch.
..!.. th =Wall thickness of head
Cs. -1-1-
(Excluding corrosion allow.)
K = Constant, see ·page 90
Q 8 = Contact angle of saddle degree
I~ 'u·z=
• ::1 § The maximum bending stress 81 may be either tension or compression .
~l!!g FORMULAS Max. Allow .. Stress Computing the tension stress in the formula for s 1, for factor K the values of
:::tlilo K 1 shall be used.
r.:i
z <"-
"'
r::lffi
Q
AT THE
SADDLES
{Tension at
QAil I 1-L+ AR'-H)
2AL
4H.
In tension S1 plus the stress due to
internal pressure (PR/2 ts) shall not
Computing the compression stress in the formula for St, for factor K the values
of Kg shall be used.
Q "'"" the Top, 1 exceed the allowable stress value of When the shell is stiffened, the value of factor K = 3.14 in the formula for S 1·
z ::tq: + 3L
s ~--+ \
Compression
>-"'
.,.;z at the shell material times the efficiency of The compression stress is not factor in a steel vessel where t/R Sio.OOS and tillL.-
=
""
..l
Q::l
"'::l
Bottom) •KR2ts
•see note on facing page
girth seam.
In compression the stress due to in-
vessel is designed to be fully stressed under internal pressure.
<
z ffi"'
u,:X: ternal pressure minus s shaRnot ex· Use stiffener ring if stress S 1 exceeds rhe maximum allowable stress.
ceed one half of the 1compression
~
Q """'
-cr::
tio AT
QL ( + 2
yield point of the rna terial or the
E
0
.,liZ)
.JO
IJJ:Z
MIDSPAN
(Tension at
R'L-; W _ 4A value given by:
z :x:- lhe Bottom 4 + 4H L
g "'"' ;::.::
~om pression
at the Top) S =+
1
3L
S, <(;)(t/R)[2 (2/3)(100)(t/R)]
If wear plate is used, in formulas for s2 for the thickness ts may be taken the
0
t 7TRzts sum of the shell and wear plate thickness, provided the wear plate extends R/1 0
inches above the horn of the saddle near the head and extends between the
~t"'l _X 2 Q ( L- 2A saddle and an adjacent stiffener ring.
tl:ii?
~1\;
IN
SHELL Sz - Rts L + 4;3 H )
!it< ;z S.2 shall not exceed 0.8 times the In unstiffened shell the maximum shear occurs at the horn of the saddle. When
c:.:: ~~ ~·
.!1 .. IN _ K 3 Q ( L- 2A ) allowable stress value of vessel rna· the head stiffness is utilized by locating the saddle close to the heads, the
< """'
,.-
... ~ SHELL Sz - Rts L + 4/3 H terial,
""
:::c:
<ll "' s3 plus stress due to internal pres·
tangential shear stress can cause an additional stress (S3) in the heads. This
stress shall be added to the stress in the heads due to internal pressure.
..l Q
< < IN S =K4Q sure shall not exceed 1.25 times the
SHELL z When stiffener rings are used, the maximum shear occurs at the equator.
i=
z "'::c0 Rts allowable tensile stress value of head
material.
""z0 !-!'::!_
""'~ IN Sz = K4Q
< "'gvN HEAD
NOTE: Use formula with factor K2 If wear plate is used, in formulas for S4 for the ~ckness ts may be taken the
!- (..)< Rth if ring not used or rings are adjacent sum of the shell and wear plate thickness and for ts may be taken the shell thick-
to the saddlq. Use formula with fac-
"'"'Q
..l ADD!-
TIONAL S =KsQ tor K3 if ring US!!d in plane of saddle.
ness squared plus the wear plate thickness squared, provided the wear plate
extends R/1 0 inches above the horn of the saddle , and A"'-R/2. The combined
Q
< STRESS 3 Rth circumferential stress at the top edge of the wear plate should also be checked.
"' IN HEAD When checking at this point: ts = shell thickness,
;::.::
Q 3K6Q =b width of saddle
00
AI!~
s.- 4ts(b+I.S~)- 2tg
f) = central angle of the wear plate but not more
ffi
AT S4 shall not exceed r.so times the than the included angle of the saddle plus 12°
~ .u. HORN allowable tensile stress value of shell If wear plate is used, in formulas for Ss for the thickness t 8 may be taken the
z ;;; ~ OF
Q
material. sum of the shell and wear nlate thickness, provided the width of the wear plate
12K6QR
""c:.::t:<.. SADDLE
s.- equals at least b + l.S6.,.f'ift;
"":::;:: ~ ~ 4ts(b+LS~) Ltg Ss shall not exceed o.s times the
compression yield point of shell rna· If the shell is not stiffened, the maximum stress occurs at the horn of the saddle.
This stress is not be to.added to the internal pressure-stress.
:;)
u AT terial.
c:.:: BOTTOM K1Q In a stiffened shell" the maximum ring-compression ls at the bottom of shell.
s Ss- Use stiffener ring if the circumferential bending stress exceeds the maximum
0 OF ts(b+ 1.56"\/Rfs)
SHELL allowable stress.
90 ,~~
91
VALUES OF CONSTANT K
(Interpolate for Intermediate Values)
o.os
m
134 0.409 0.971 0.667 0.347 0.714 0.722
136 0.420 0.946 0.641 0.340 0.708 0.740 >
138 0.432 0.923 0.616 0.334 0.702 0.759
140 0.443 0.900 0.319 0.592 0.327 0.697 0.780 0.02
142 0.455 0.879 For 0.569 0.320 See 0.692 0.796
144 0.467 0.858 Any 0.547 0.314 chart 0.687 0.813
146 0.480 0.837 Con- 0.526 0.308 on 0.682 0.831
148 0.492 0.818 Tact 0.505 0.301 facing 0.678 0.853
150 0.505 0.799 Angles 0.485 0.295 0.673 0.876
page
152 0.518 0.781 6 0.466 0.289 0.669 0.894
154 0.531 0.763 0.448 0.283 0.665 0.913
156 0.544 0.746 0.430 0.278 0.661 0.933
158 0.557 0.729 0.413 0.272 0.657 0.954
160 0.571 0.713 0.396 0.266 0.654 0.976
162 0.585 0.698 0.380 0.261 0.650 0.994
164 0.599 0.683 0.365 0.256 0.647 1.013
166 0.613 0.668 0.350 0.250 0.643 1.033
168 0.627 0.654 0.336 0.245 0.640 1.054
170 0.642 0.640 0.322 0.240 0.637 1.079 RATIOA/R
172 0.657 0.627 0.309 0.235 0.635 1.097
174 0.672 0.614 0.296 0.230 0.632 1.116
176 0.687 0.601 0.283 0.225 0.629 1.137
178 0.702 0.589 0.271 0.220 0.627 1.158
180 0.718 0.577 0.260 0.216 0.624 1.183 /
92 93
STRESSES IN LARGE HORIZONTAL VESSELS SUPPORTED BY 1WO STRESSES IN LARGE HORIZONTAL VESSELS SUPPORTED BY 1WO
SADDLES SADDLES
EXAMPLE CALCULATIONS EXAMPLE CALCULATIONS (cont.)
Design Data
A = 48 in. distance from tangent line TANGENTIAL SHEAR STRESS (S,)
of head to the center of saddle
· b = 24 in. width of saddle Since A (48)>R/2 ((i0/2), the applicable formula:
H = 21 in. depth of dish of head
s2 = K2 Q(_].- 2A ) = 1.171 X 300,000 ( 960 - 2 X 48 ) = 5,120psi
L = 960 in. length of vessel tan.-tan.
Rt, \I+4f3H 60 X 1 960 + 4/3 X 21
P = 250 psi. internal design pressure
Q = 300,000 lb. load on one saddle s 2 does not exceed the stress value of shell material multiplied by 0.8; 20,000 X 0.8
R = 60 in. outside radius of shell
= 16,000 psi
ts = 1.00 in. thickness of shell
9 = 120 deg. contact angle
Shell material: SA 515-70 plate
Allowable stress value 17,500 psi. CIRCUMFERENTIAL STRESS
Yield point 38,000 psi. I
Joint Efficiency: 0.85 Stress at the horn of saddle (S4J
Since L (900) > 8R(480), A(48) > R/2 (60/2), the applicable formula:
S=- Q JK,Q
4 4t5 (b+l.56 ./Rt'sJ - :zrf
LONGITUDINAL BENDING STRESS (S,)
AIR = 48/60 = 0.8; K = 0.036 (from chart)
Stress at the saddles 300,000 3 X 0.036 X 300,000
t • = 20,000 psi
54 =- 4 x. 1 (24 + 1.56 v 60 x 1) 2
~
2 2
- + __..:.:=,......--=;;...._
I - 48 60 -21 S 4 does not exceed the stress value of shell material multiplied by 1.5; 20,000 X 1.5
300,()()() X 48 - 960 2 X 48 X 960
= 30,000 psi
( 1 +-4_x_21_
3x 960 . Stress at bottom of shell (5 5 )
0.335 X 60 2 X 1 = 522 psi.
K7Q
Ss =- --~=---==
t1 (b + 1.56 yRtsJ
.QL(
4
2
+ R i_2 H
1 4H L
4A)
300,000x960
4 4 X 21
_ 4x 48)
960
Ss does not exceed the C()mpression yield point multiplied by 0.5; 38,000 x 0.5
1+ = 19,000 psi
+ 3L .
sl = -----'--~=----.;__ =----::--~~3:.,;x,:....:.960::__ ____!__ =
rc Rl t, - 3.14 x 602 x 1 4959 psi
Compression stress is not a factor since t!R > 0.005; 1/60 = 0.017
94 95
,--------------------, -----..... __
STIFFENER RING
FOR LARGE HORIZONTAL VESSELS SUPPORTED BY STIFFENER RING
SADDLES FOR LARGE HORIZONTAL VESSELS
SUPPORTED BY SADDLES
NOTATION.
A = Cross sectional area of ring plus the
effective area of shell, in2 VALUES OF CONSTANT, X
1 = Moment of inertia, in4
K = Constant, see next page (Interpolate for Intermediate Values)
Q = Load on one saddle, lbs.
R = Radius of shell, in. Contact
S6 = Max. combined stress, psi.
Angle e 150° 160°
(J Contact angle, degree
Max. .34 .33 .32 .30 .29
TYPE OF RING MAX. STRESS FORMULAS Allow
Stress .053 .045 .037 .032 .026
'L Saddle Tip of the 6 A 1/d .:a.., ..C::~ CALCULATION OF MOMENT OF INTERIA (!)
and Ring ~ lr
J
........;:.r Ring
<I) :ao 1. Determine the width of shell that is effective to resist the circumferential bend-
~ ii:. Saddle Ring Inside. ] ~ ing moment The effective width 1.56~; 0. 78~ on both sides of stiffener
andRmg ~~ooo Compression s6=_K9Q_KloQR -~-a ring .
. . :~==~~~~~-~ ~~-~a~t=the~S=he~l~l~~---------A-----I/_c
•
7
Governs
Ring Outside.
____ ~ !~
"" :;;
·;:;: E
2. Divide the stiffener ring into rectangles and calculate the areas (a) of each
tr+J.s6v'Rii.
I • .. j Stress at the
Shell
S6=_K9Q + ~~ oOR
A 1/c
;
c ...
.!lf rectangle, including the area of shell connection within the effective width.
Add the areas (a) total area A.
D I fi.Saddle Ring Inside. .§~ 3. Multiply the areas (a) with the distances (Y) from the shell to the center of
..;;;;..~ andRing 1--f+-.'r Stressatthe S6 =_K9Q+K1oOR ,.. cu
~ J~
<
2(1r+I.SovRts)
rr I Ifl Tip
::~Inside.
Stress
of at
thethe
A
K Q K QR
S6=--L- .......!Jt_
l/c i~
...... 5
0 0
4.
gravity 9f the rectangles. Summarize the results and denote allAY.
Determine the neutral axis of the stiffening ring, the distance (C) from the shell
to the neutral axis C ~y
.. t- •. A 1/d cu o.
~,----------------r~ru=·n~g~------~------------------~ ~]
'El =~ 5. Determine the distances (h) from the neutral axis to the center of gravity of
i-:::.J ~ >,
.. ..I
2 (tr+ J.S6y'Rii)
each rectangle of the stiffener.
Ring Outsi~e. ~~
{\. I Compression s6 =-K9 Q - K I 0 QR Multiply the square of distances (h 2) by the areas (a) and summarize the
-Tj·r.f. Lt;
v• v•
6.
::dl• ~~~~;shell A 1/c ~] results to obtain AH2 :
J
7. , where b = the
~
I \_Saddle §
Ring lnside. : Width and d = the depth of the rectangles.
1-- and Ring
d
Stress at the
Shell
s6 =- K9Q + Kl oOR
. A 1/ c
.;
§~
·a 8. The sum of AH2 and £ lg gives the moment of intertia of the stiffener ring and
I
· - c r-t>:~~~----~----------------~
Ring Inside. ·~
.: Eo. the effective area of the shell.
2 .. ~
(tr+I.S~ • ...
I Stressatthe
Tip of the
S6=_K9Q_KioQR
A 1/d
e 0
..5 ..5
0
see example calculatio~s on the following pages.
Ring
96
97
STIFFENING RINGS STIFFENING RINGS
Moment oflnertia (I)- Example Calculations
Moment oflnertia (/)-Example Calculations
(All dimensions in inches- R 72 in. outside radius of shell)
(All dimensions in inches R 72 in. outside radius ofshell)
I 0.781Rd;_ Cj b3=4.00 .J I = 0.78 .fifdi =
o. 78 vr-:72::--x-:::o-=.5 = 4.68 .;. •;g ·~t;;V7,;~~'7n3f_;//7;/~AL----.-- 1--o.:..:..7~8::--l/~!72~__x_ol_
. 5_4._6s_ _,
~
..,; <L Saddle "::::: ~ AREA W t'l\ Ig
AREA(Dig 11 8. and,~Rin~"g&.--fl!V1 If_:;: -<::1f. b J: 9 86 0 53
_j_J__ • X • = 0 103 . 4
b1df 9.86 X 0.53 ~ "\:! 'IE, - :cior. 12 - 12 · m.
0.103 in. 4
12 12 "'-:I-"\:! - - X ~ X ~ ~ ~ AREAQ}Ig
~ SHELL-,. ~ ~ b2t/j _ 0.5 X 63 _ OO . 4
AREAQ)Ig It "II $1::/"////(I~// --u-----rz- · 9 m.
~
MARK
OF AREA y MARKOF AREA y h1 ax h2 bJE_
AREAS a axy h h2 axh2 AREAS a axy h
12
m
1 4.93 0.25 1.23 1.23 1.51 7.44 l . 4.93 0.25 1.23 2.29 5.24 25.83 0.10
2 3.00 3.50 10.50 2.02 4.08 12.24 9.00 3.50 10.50 0.96 0.92 2.76 9.00
17.72 35.44 0.04
TOTAL A=7.93 - AY=11.73 - - AH2=19 lg=9.10
- AH=64.03 IJF9.14
C == AY = 11.73= 148 I= AJf2 + Ig == 19.68 + 9.10 = 28.78 in.4 C AY = 25.23 = 254
A 7.93 . I= Aif2 + lg = 64.03 + 9.14 = 73.17 in.4
A 9.93 .
1-1.56 Wid;
1.56 ...Jn x 0.25 =6.618
AREA(Dig
b1ai _ 13.74 x 0.253 _ " m·
12- 12 - 0.
02 .4
..... .....
AREAQ)
2b2~ = 0.50 x 63 = 9 .00 in.4
<'1
0 · b 2+1=6"Moi:S
bJ-13.74
b2+I=6.8o8 -
<'1
II
;.;::
12 12
MARKS AREA bel MARKS AREA bel
OFAREAS. a !? axh2 12 OF AREAS a y axy h !? axh2 12
3.43 1 3.4J 0..125 0.43 2.59 6.72 23.09 0.02
0.43 2.12 7.27 0,02 ~
,.
i 2 3.00 3.250 9.75 0.53 0.28 0.84 9.00
3.00 3.250 9.75 1.670 2.79 8.37 9.00
3 2.00 6.375 12.75 3.66 13.40 26.80 O.oi
TOTAL A=6.4 AY= 10.18 - AH2 = 15.64 Ig= 9.02 TOTAL A 8.43 AY= 22.93 - - 1
AH -50.73 /g-9.03
H0RN6F/~~ ~ l!J
WEAR
PLATE
I t;:;l ~
SADDLE I
v-~
I
I MAX.
~/EFFECTIVE
f"""'"l'-·
~.~ AREA
I , \.DOLTS
~ 2 .I 2
I : \. DOLTS : I I q_ SA,DDLES
!
I. The saddle at the lowest section must resist the horizontal force (F). The effective
cross section of the saddle to resist this load is one third of the vessel radius (R).
F=K11 Q, Where Q= the load on one saddle, lbs.
K11 =constant as tabulated.
-t--+ +-· '"'
o;;r
The average stress shall not exceed two thirds ofthe compression yield point ofthe EXPANDING VESSEL CONTRACTING VESSEL
material. (See example below.)
For thermal expansion and contraction, one of the saddles, preferably the one
VALUES OF CONSTANT K11 on the opposite side of the pipe connections, must be allowed to move. In this
Contact Angle I 120° I 130° 140° I 150° 160° I 170° I 180° saddle for the anchor bolts slots are to be used instead of holes. The length of
.241 I .259 the slots shall be determined by the expected magnitude of the movement. The
Kn I .204 I .222 .279 I .298 I .318
coefficient of linear expansion for carbon steel per unit length and per degree
F =0.0000067. The table below shows the minimum length of the slot. Dimen-
EXAMPLE: sion "a" calculated for the linear expansion of carbon steel material between 700F
Diameter of vessel= 8'- 6" and the indicated temperature. When the change in the distance between the saddles
Weight of vessel= 375,000 lbs. is more than 3/8" inch long, a slide (bearing) plate should be used. When the
vessel is supported by concrete saddles, an elastic, waterproof sheet at least 1/4"
Q = 187,500 lbs. thick is to be applied between the shell and the saddle.
Saddle material: SA 285 C
Web plate thickness= 0.25 in.
Contact angle= 120° MINIMUM LENGTH OF SLOT (DIM. "a")
K 11 = 0.204 from table above
R/3 = 5113 = 17 inches DISTANCE FOR TEMPERATURE OF
E~D
BETWEEN
Force, F = K 11 xQ = 0.204 x 187,500 = 38,250 lb. . SADDLES
-50 100 200 300 600 700 800 900
To resist this force the effective area of web plate= R/3 x 0.25 =4.25 in. 2 Ft. 400 500
38,250/4.25 = 9,000 lbs. per square inch. 10 0 0 0 1/4 3/8 3/8 1/2 5/8 3/4 3/4
0 0 1/4 3/8 5/8 3/4 1 1-1/8 1-1/4 1-3/8
The allowable stress = % x 30,000 = 20,000 psi. ., 0
.. 20
. 30 - 1/8 3/8 5/8 1-3/8 1-5/8 1-5/8 2
1/4 7/8 1-1/8
The thickness of the web plate is satisfactory for horizofital force (F). . .,..
'5
'0
00
iii
c 4Q 1/4 1/8 3/8 3/4 1-1/8 1-1/2 1-7/8 2-1/8 2-3/8 2-1/2
50 3/8 1/4 1/2 1 1-3/8 1-5/8 2-1/4 2-5/8 3 3-3/8
2. The base plate and wear plate should be thick enough to resist longitudi- 60 3/8 1/4 5/8 1-1/4 1-5/8 2-1/8 2-3/4 3-l/8 3-5/8 4-1/8
nal bending over the web.
""' of
The width 70 1/2 1/4 3/4 1-3/8 1-7/8 2-1/2 3-1/8 3-5/8 4-1/4 4-5/8
the slot equals 80 3/8 3/4 1-1/2 2-1/8 2-7/8 3-5/8 4-1/8 4-7/8 5-3/8
1/2
3. The web plate should be stiffened with ribs against the buckling. the diam. of 3-1/4 4 4-5/8 5-3/8 6
90 5/8 3/8 7/8 1-3/4 2-3/8
anchor bolt +
100 5/8 3/8 1 1-7/8 2-5/8 3-5/8 4-1/2 5-1/8 6 6-5/8
".
~"
100 101
SADDLE
FOR SUPPORT OF HORIZONTAL VESSELS SADDLE
NOMINAL PLATETIIICKNESS
DIMENSIONS INCHES
DIAM. NO. MAXIMUM
OF BOLT OF WEB, WEIGHT
VESSEL
A B c D E
DIAM. RIBS BASE
FLANGE,
WEAR ON VESSEL
FT.-IN. FT.- IN. IN. IN. FT.-IN. G K
FT. ·IN. INCH RIBSH
~
10-6 9-1 Y:. 6-0 9 24 3-6 1Y. 2 1 Y. Y:.
11-0 9-6Y:. 6-3 9 24 3-8 1Y. 2 I % 'lz
11-6 10-0 6-6 9. 24 3-10 1Y. 3 1 Y. 'lz 00
SEE FACING PAGE FOR DIMENSIONS
12-0 10-5 6-9 9 24 4-0 1'!. 3 1 Y. 'lz 1076000
102
103
STRESSES IN VESSELS ON
STRESSES IN VESSELS ON LEG SUPPORT
LEG SUPPORT
NOTATION:
W = Weight of vessel, pounds
0.30
n = Number oflegs \ I
Q
R
= W Load on one leg, pounds
n
= Radius of head, inch
0.25
1\.
1\ __.--K,
+-
H = Lever arm ofload, inch
2A, 2B = Dimension ofwearplate
s ~0.15
0.20
, ......
_,.. v
~
~
0.25
0.20
'1\
\
-+-
Positive values denote tensile stresses and negative values denote compression.
Computing the maximum tensile stresses, in formulas for S , S and K , K , K
1 2 1 3 5
~ 0.15 \
l
, _......- /
K2
vK6
and K7 denote negative factors and K2, K4, K6 and K denote positive factors. \. /
8 0.10 /
Computing the maximum compression stresses in formulas for S , S and K , K ,
K3, K4, K5. Ka- K7 and K8 denote negative factors. 1 2 1 2 f\ k .......
0.05 ""-..
The maximum tensile stresses S1, and S2, respectively, plus the tensile stress due
to internal pressure shall not exceed the allowable tensile stress value of head
material. f'l ~ \0 QO 0
" r-
f'l
1--.
II") ~
I
oooo.....:~.....: N
. VALUE OF K2 , & K6
D
The maximum compression stresses S1• and S2, respectively, plus the tensile stress
due to internal pressure shall not exceed the allowable compression stress value of
head material.
105
104
STRESSES IN VESSELS ON LEG SUPPORT
EXAMPLE CALCULATIONS
STRESSES IN VESSELS ON LEG SUPPORT
DESIGN DATA
W = 800,000 lb. weight of vessel
n = 4, number of legs
w 800,000 = 200,000 lb. load on one leg
Q = --;; = - -
4
R = 100 inch, radius of head
0.20Ht-t---t-H-t--t--+--l--1---l---l H = 5 inch, lever arm ofload
/1--1'\
2A = 30 inch, 2B = 30 inch, dimensions of wear plate
t<t-.. 0.15 I'\ /Kj t = 1.8 i~ch thickness of head
~ rrt-rrT~,rh(~~~~--~-+~ cos oc = 0.800
t< 0.1 Ofi7Tt-t-+-1~~,1-'-..,--+---l----l-~f---l P = 100 psi, internal pressure
Head material: SA 515-70
"t'-..... Allowable stress value: 20,000 psi
0.05
_L
v ........ ".
K7
1-- -.........r--
..........._
Joint Efficiency: 0.85
Yield Point: 3 8,000 psi
0~~~~~-+--~r-----~r~-=+==~--~~
Factors K (see charts):
C'l ~
1.0 00 0 C'l V) 0 C= -fAR= "'h5 x 15 = 15 inch
cicicici....;....; .....; ('(') D
D = 1.82 ~ # = 1.82 l~O -f1f= 2.03
LONGITUDINAL STRES:
0.60 "liit-t---t-H-t--t--+--1---1---1--.:....l 1.) Maximum tensile stress:
0 0
The sum of tensional .stresses:
C'l ('(') 7.634 + 2.778 = 10,412 psi
VALUE OF K4 , & Ka
D
It does not exceed the stress value of the girth seam:
20,000 X 0.85 = 17,000,
106 107
-
1 2 4
Circumferential stress:
1.) Maximum tensile stress:
82 =~ [coscc(-Kj+6K6 )+ ~~ ~(-K--6K11)]
- 200,000 [ 5 - ,-
S]- 1.82 0.800(-0.020+6x0.010)+ 100 \~~ (-0.022+6xO.OlO)]
-v
= + 2,849 psi
The stress due to internal pressure:
PR 100 X 100
2t = 2 x 1.8 + 2,778 psi
The sum of tensile stresses:
-2,849+2,778 5,627psi
It does not exceed the stress value of the girth seam:
20,000x0.85=17,000psi ·
~
r R ----!
!
--tt- I
r 28 "I
Notch out angles
2If]Jl
to clear seam
Uli i
~ J.:----,+----.:t~
~ _ulY.l
Q \ Q
1 I
UN STIFFENED STIFFENED
SHELL SHELL
-C'J NOTATION:
W = Weight of vessel, lb
n = Number of lugs
2A, 2B = Dimensions of wear plate
S = Stress, pound per sq. in
t = Wall thickness of shell, in
SECfiON A-A C = shape factor, see table
Q = W = Load on one lug, lb K = Factors, see charts
n
R = Radius of shell, in D =d. ,..3(T"
H = Lever ann of load. in R VA
LONGITUDINAL STRESS:
QH
s1 = -+ --
DRt
2
VESSEL VESSEL ANGLE 1 w
DIA HEIGHT MAX SIZE max
NOfE: In tensionS 1 plus the stress due to internal pressure PR12t shall not exceed
2'-6"
3'-0" 4"
the stress value of shell material times the efficiency of girth seam.
8'-0" 3"x3"x3/8"
3'-6"
5'-0"
4'-0"
10'-0" 3.5" X 3.5" X 3/8" 6"
4'-6" CIRCUMFERENTIAL STRESS:
5'-0"
14'-0" 4"X4"X l/2" 7"
5'-6" s2 = ± QH,( K4 R)
c3 K3 + 6 -c
6'-0" DR 2 t 4t
16'-0" 5"x5"x 1/2" 10"
6'-6" 7'-0" NOIE: In tension S2 plus the stress due to internal pressure PR/t shall not exceed
7'-0"
18'-0" 6"x6"x5/8" 1'-0" the stress value of shell material multiplied by 1.5.
7'-6"