Welded Tanks for Oil StorageP-1

P.3.12 Sample Problem No. 3

This sample problem uses the alternative method on the data presented in the sample problem in P.2.9. The calculated limit loads from in P.2.9 are assumed to be from thermal loads. The stresses are calculated using the equations from Tables P-2 through P-4. Stress factors are calculated by interpolating between t/tn = 2 and 1 to get the stress factor for t/tn = 1.092 at d/tn = 30 , then again at d/tn = 10, and finally by interpolating between d/tn = 30 and 10 to get the stress factor for d/tn = 19.70.

P.3.12.1 Data

Tank diameterD=80 m (260 ft or 3120 in)

Tank shell thicknesst=34 mm (1.33 in)

Nozzle outside diameterd=610 mm (24 in)

Nozzle neck thickness (assumed)tn=31 mm (1.218 in)

Nozzle location from bottomL=630 mm (24.75 in)

Design stressSd=160 MPa (23,200 lbf/in.2)

Using the limit loads from P.2.9.3:

FR =320328,000 N (74800 lbs)

MC =550  106 Nmm (4.95  106 inlbfs)

ML =195  106 Nmm (1.74  106 inlbfs)

MT =VC = VL = 0

P.3.12.2 Solution

P.3.12.2.1 Establish the values for t, u, d/tn, and t/tn from the data provided.

In SI units:

u=(d/D)(D/t)0.5

u=(610/80,000)(80,000/34)0.5

=0.37

d/tn=610/31

=2019.7

t/tn=34/31

=1.096

Use t/tn=1.096

In US Customary units:

u=(d/D)(D/t)0.5

u=(24/3120)(3120/1.33)0.5

=0.37

d/tn=24/1.218

=2019.7

t/tn=1.33/1.218

=1.092

Use t/tn=1.092

P.3.12.2.2 Bending and membrane stress factors for FR, MC and ML are determined by interpolation and

are reported in Table P-7. Values in Table P-7 are shown to four places passed thr decimal point to

help verify computer-generated data.

EDITOR’S NOTE:Please change existing “Table P-7” to be “Table P-8”. Also change reference to “P-7”, that appears in section P.3.11.2.4 , to refer to “P-8”.

Table P-7 --- Stress Factors for Sample No. 3 in U.S. Customary units

value / d/tn / t/tn / Use eqn # / factor / interpolate / interpolate
fr for FR / 2 / 17 / 1.6536
& d/tn=30 / 1.092 / 2.1822
1 / 16 / 2.2357
19.70 / 2.1577
fr for FR / 2 / 13 / 1.5854
& d/tn=10 / 1.092 / 2.1347
1 / 12 / 2.1903
ffor FR / 2 / 36 / 1.2118
& d/tn=30 / 1.092 / 0.9966
1 / 35 / 0.9748
19.70 / 0.9384
ffor FR / 2 / 32 / 0.9960
& d/tn=10 / 1.092 / 0.8836
1 / 31 / 0.8722
fr for MC / 2 / 55 / 1.5511
& d/tn=30 / 1.092 / 1.8360
1 / 54 / 1.8648
19.70 / 1.8538
fr for MC / 2 / 51 / 1.5688
& d/tn=10 / 1.092 / 1.8706
1 / 50 / 1.9011
ffor MC / 2 / 74 / 1.1410
& d/tn=30 / 1.092 / 0.7792
1 / 73 / 0.7426
19.70 / 0.8165
ffor MC / 2 / 70 / 1.2371
& d/tn=10 / 1.092 / 0.8516
1 / 69 / 0.8125
fr for ML / 2 / 93 / 1.5006
& d/tn=30 / 1.092 / 1.7320
1 / 92 / 1.7554
19.70 / 1.7305
fr for ML / 2 / 89 / 1.5111
& d/tn=10 / 1.092 / 1.7305
1 / 88 / 1.7527
ffor ML / 2 / 112 / 1.0592
& d/tn=30 / 1.092 / 0.7359
1 / 111 / 0.7031
19.70 / 0.7218
ffor ML / 2 / 108 / 1.0733
& d/tn=10 / 1.092 / 0.7085
1 / 107 / 0.6716

P.3.12.2.2 To calculate the stress factors, interpolation is required between d/t = 10 and 30 to arrive at the values for d/t = 18

FRfT for d/tn= 10 and t/tn = 1 (12)

fr=–0.9384ln(u) + 1.2638

=–0.9384ln(0.37) + 1.2638

=2.197

fT for d/tn= 30 and t/tn = 1 (16)

fr=–0.9074ln(u) + 1.3398

=–0.9074ln(0.37) + 1.3398

=2.242

fT for d/tn= 20 and t/tn = 1

fr=0.5(2.197 + 2.242)

=2.219

FRf for d/tn= 20 and t/tn = 1 (31 and 35)

f=0.5{(–0.3427ln(u) + 0.5338) + (–0.344ln(u) + 0.6352)}

=0.9259

MCfr for d/tn= 20 and t/tn = 1 (50 and 54)

fr= 0.5{(–0.0233u2 – 0.1xu + 1.9416) + (–0.0207u2 – 0.0936xu + 1.9026)}

= 1.883

f= 0.5{(0.0229u2 – 0.1966u + 0.8826) + (0.0048u2 – 0.0649u + 0.7661)}(69 and 73)

=1.322

MLf= 0.5{(0.0769u2 – 0.42u + 0.8174) + (0.0205u2 – 0.2132u + 0.7797)}(107 and 111)

=0.688

fr=0.5{(–0.0359u2 – 0.5507u + 1.9629) + (0.0658u2 – 0.695u + 2.0052)}(88 and 92)

=1.7556

P.3.12.2.3 Calculate the stresses.

In SI units:

Stress due to FR,

r=(FR/t2)fr

r=(328,000/342)2.2192.1577

=629 612 MPa

=(FR/t2)f

=(328,000  0.9259/342 ) x 0.9384

=263 266 MPa

Stress due to MC,

r=(MC/(dxt2)fr

r=(550  106)/(610  342)x1.8831.8538

=1469 1446 MPa

=(MC/(dt2)f

=(550  106)/(610 342)1.3220.8165

=1031 637 MPa

Stress due to ML,

r=(ML/(dt2)fr

r=(195  106)/(610  342)1.75561.7305

=485 479 MPa

=(ML/(dt2)f

=(195  106)/(610  342)0.6880.7218

=190 200 MPa

In US Customary units:

Stress due to FR,

r=(FR/t2)fr

=(74,800/1.332)2.2192.1577

=93,83391,240 lbf/in.2

=(FR/t2)f

=(74,800  0.9259/1.332  0.9384

=39,15339,682 lbf/in.2

Stress due to MC,

r=(MC/(dt2)fr

=(4.95  106)/(24  1.332)1.8831.8538

=199,594216,148 lbf/in.2MPa

=(MC /(dt2)f

=(4.95  106)/(24  1.332)1.3220.8165

=140,12995,198 lbf/in.2

Stress due to ML,

r=(ML/(dt2)xfr

=(1.74  106)/(24  1.332)1.75561.7305

=71,95570,956 lbf/in.2

=(ML/(dt2)f

=(1.74  106)/(24  1.332)0.6880.7218

=28,19829,583 lbf/in.2

P.3.12.2.4 Calculate the stress reduction factors.

In SI units:

B=2(Dt)0.5

B=2(80,000  34)0.5

=3298 mm

h=L/B

h=630/3298

=0.19

In US Customary units:

B=2(Dt)0.5

=2(3120  1.33)0.5

=129 in.

h=L/B

=24.75/129

=0.19

P.3.12.2.5 Calculate the stress intensity.

S = 0.5z[(r + ) ± {(r – )2 + 42}0.5]

Where z = 0.470.46 from Figure P-11, and Smax is from the combination of FR and MC.

In SI units:

SmaxS=z x Smax = 0.46 x 0.5  0.47[(629 + 263 2058+903) ± {(2058-903 629 – 263)2 + 402}0.5]

=296 946 MPa

In US Customary units:

SmaxS=z x Smax = 0.46 x 0.5  0.47[(93833 + 39153307,388+134,880) ± {(307,388-134,880 93833 – 39153)2 + 402}0.5]

=44,102 141,398 lbf/in.2

For this sample problem it is assumed that the limit loads include the liquid load from the tank, therefore the preceding stresses are total stresses. In this case the allowable stress is:

Sall =2.0 SdMechanical Load

=3.02.5SdThermal Load

In this problem it is assumed that the loads are of thermal nature, therefore the allowable stresses are:

In SI units:

Sall =3.02.5 160 MPa

=480 400 MPa

Sall Smax Sall

480 MPa296 946 MPa > 400 MPa

In US Customary units:

Sall =3.02.5x 23,200 lbf/in.2

=69,60058,000 lbf/in.2

Smax Sall SallSmax

69,600141,398 lbf/in.2 44,102 58,000 lbf/in.2

P.3.12.3 Conclusion

Based on the assumptions made, this analysis indicates that the piping arrangement for example P.2.9 is NOT acceptable.

EDITOR’S NOTE:ALL FIGURES ARE UNCHANGED, EXCEPT : IN THE UPPER LEFT CORNER OF FIGURE P-9B, CHANGE “d/tn=3” TO READ “d/tn=30”

FigureP-8A—Stress Factor fR Due to Radial Thrust FR, d/tn = 10
FigureP-8B—Stress Factor fR Due to Radial Thrust FR, d/tn = 30
FigureP-8C—Stress Factor fR Due to Radial Thrust FR, d/tn = 50
FigureP-8D—Stress Factor fR Due to Radial Thrust FR, d/tn = 100
FigureP-8E—Stress Factor f Due to Radial Thrust FR, d/tn = 10
FigureP-8F—Stress Factor f Due to Radial Thrust FR, d/tn = 30
FigureP-8G—Stress Factor f Due to Radial Thrust FR, d/tn = 50
FigureP-8H—Stress Factor f Due to Radial Thrust FR, d/tn = 100
FigureP-9A—Stress Factor fr Due to Circumferential Moment MC, d/tn = 10
FigureP-9B—Stress Factor fr Due to Circumferential Moment MC, d/tn = 30
FigureP-9C—Stress Factor fr Due to Circumferential Moment MC, d/tn = 50
FigureP-9D—Stress Factor fr Due to Circumferential Moment MC, d/tn = 100
FigureP-9E—Stress Factor f Due to Circumferential Moment MC, d/tn = 10
FigureP-9F—Stress Factor f Due to Circumferential Moment MC, d/tn = 30
FigureP-9G—Stress Factor f Due to Circumferential Moment MC, d/tn = 50
FigureP-9H—Stress Factor f Due to Circumferential Moment MC, d/tn = 100
FigureP-10A—Stress Factor fr Due to Longitudinal Moment ML, d/tn = 10
FigureP-10B—Stress Factor fr Due to Longitudinal Moment ML, d/tn = 30
FigureP-10C—Stress Factor fr Due to Longitudinal Moment ML, d/tn = 50
FigureP-10D—Stress Factor fr Due to Longitudinal Moment ML, d/tn = 100
FigureP-10E—Stress Factor f Due to Longitudinal Moment ML, d/tn = 10
FigureP-10F—Stress Factor f Due to Longitudinal Moment ML, d/tn = 30
FigureP-10G—Stress Factor f Due to Longitudinal Moment ML, d/tn = 50
FigureP-10H—Stress Factor f Due to Longitudinal Moment ML, d/tn = 100
FigureP-11—Stress Reduction Factor