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7/30/2019 Mathcad - Example 8
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Example problem 6.3 from API 579 example manual Darren Ting April 21 2013
Material is SA 516 Grade 70 year 1980
P 500psiDesign Pressure
Tdesign 450 FDesign Temp
D 60inInside Diameter
Tnom 1.125inWall Thickness
LOSS 0.03inUnifrom Metal Loss
FCA 0.05inFuture Corrosion ALlowanceSA 17500psi
Allowable Stress
Weld Joint Efficiency(Long and Circ) E_l 0.85 E_c 0.85
Saddle Reaction Force Q_s 34690lbf
Mid Span Bending Moment M 1312600in lbf
Tangent to Tangent Length L_l 30ft
Depth of Head L_h 15in
DIstance from support to tangent L_a 4ft
Applied net section shear force for weight/ plus thermal V_l 0 lbf
Applied net section torsion fr weight/plus thermal M_t 0lbf in
Applied section axial force for weight/plus thermal F_axial 0lbf
Assuming
RSF_a 0.9
0
Refer to 6.2 of API 579 for and limitations of procedure. All these limitations are assumed
for this analysis.
7/30/2019 Mathcad - Example 8
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Inspection data in vector form
Pit_Couple
1
2
3
4
5
6
7
8
9
0
11
12
13
14
15
16
17
18
19
P_k
3.5
4.2
2.7
2.1
4.6
3.1
2.9
3.1
2.6
2.2
11.8
2.5
3.8
1.9
1.8
1
2.5
1.5
1.3
in _k
10
15
22
30
5
15
20
45
60
0
10
20
35
90
0
22
45
67
90
d_ik
0.5
1.6
0.9
1
.7
1.1
.8
.5
1.3
.4
1.5
.6
2.4
.4
1
.6
.9
.6
.8
in w_ik
.5
.6
.5
.7
.6
.5
.65
.4
.5
.55
.4
.75
.5
.25
.7
.75
.3
.5
.4
in
d_jk
0.6
1.8
.9
1.2
1.2
2.2
.5
1
.8
.3
.8
.5
1.6
.8
.8
.2
1.2
.6
.5
in w_jk
.4
.65
.75
.6
.5
.45
.6
.75
.2
.75
.5
.7
.75
.5
.5
.7
.4
.7
.7
in
Perform level 2 Assessment as per par. 6.4.3
7/30/2019 Mathcad - Example 8
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Uniform Thickness away from pitted region established by thickness measurement,T_rd
T_rd Tnom LOSS 1.095 in
D_o D 2Tnom 62.25 in
R_c1
2D LOSS FCA 30.08 in
Future corroded thickenss T_c T_rd FCA 1.045 in
From 6.4.3.2 to establish MAWP for regoin with pitting damage
Calculate average pit depth
W_avg_1w_ik w_jk( )
2
W_avg_1
0.45
0.625
0.625
0.65
0.55
0.475
0.625
0.575
0.35
0.65
0.45
0.725
0.625
0.375
0.6
0.725
0.35
0.6
0.55
in
7/30/2019 Mathcad - Example 8
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R_m
R_cD_o
2
230.6025 in
From Annex A .7.3 API 579(2009)-Note that 2007 equation is slighty different
T_sl3 Q_s L_l( )
SA E_l R_m( )2
12 R_m( )
2L_h
2
L_l( )2
14 L_h( )
3 L_l
4L_a
L_l
0.3633 in
Check SA 1.75 104 psi Q_s 3.469 104 lbf
E_l 0.85 L_l 30 f t
R_m 30.6025 in L_h 15 in
_1P
E_c
0.6R_c
T_c
1.7285 104
psi
_2P
2E_l
0.4R_c
T_c T_sl
1.2861 104
psi
7/30/2019 Mathcad - Example 8
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Check circumferential stress condition
Statement 1
P 500 p siSA E_c 0.385 5.7269 10
3 psi true "true" false "false"
result_1 if P SA E_c 0.385 true false( )
result_1 "true"
Hence part a) is used
Check longitudinal stress condition
Statement 1
P 500 p si
SA E_l 0.385 5.7269 103
psi
result_2 if P SA E_l 0.385 true false( )
result_2 "true"
Hence part c) is used
7/30/2019 Mathcad - Example 8
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MAWP_cSA E_c T_c( )
R_c 0.6T_c506.216 psi
MAWP_l2SA T_c T_sl( ) E_c
R_c 0.4 T_c T_sl( )680.3423 psi
Governing MAWP
MAWP min MAWP_c MAWP_l( ) 506.216 psi
7/30/2019 Mathcad - Example 8
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Calculate average pit depth Calculate avg1
D_avg_kd_ik d_jk( )
2
0.55
1.7
0.9
1.1
0.95
1.65
0.65
0.75
1.05
0.35
1.15
0.55
2
0.6
0.9
0.4
1.05
0.6
0.65
in_avg1
P_k D_avg_k( )
P_k
0.8429
0.5952
0.6667
0.4762
0.7935
0.4677
0.7759
0.7581
0.5962
0.8409
0.9025
0.78
0.4737
0.6842
0.5
0.6
0.58
0.6
0.5
Calculate _21Calculate _11
_21_2
_avg1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
41.5259019710
42.1606771910
41.9291760610
42.7008464910
41.6208602510
42.7496302510
41.6576623910
41.6965803710
42.1573581810
41.5294368810
41.424993910
41.6488684310
42.7151366810
41.8797100110
42.5722347510
...
psi_11 _1
_avg1
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
42.050775910
42.9038986810
42.5927666810
43.6298733510
42.1783975710
43.6954375610
42.2278587710
42.280163610
42.89943810
42.0555267310
41.9151578610
42.2160398910
43.6490790310
42.5262854810
43.4570222310
...
psi
7/30/2019 Mathcad - Example 8
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_1 _112 cos _k( )( )4 sin 2_k( )( )2
3 sin 2_k( )( )
2_11 _21( )
2 sin _k( )( )
4sin 2_k( )( )
2 _21( )
2
Note typo in api 2007 example for_1. There should not be
numerical values in given formula
Check _110
2.0507759 104
psi
cos _k( )0
0.9848
sin 2_k( )0
0.342
_210
1.52590197 104
psi
_1
84.1732440210
88.2840514410
86.4607170310
91.219111710
84.7363555710
91.3415646510
84.8049392910
84.2941925410
85.902089810
84.2251901510
83.6395402110
84.7540937310
91.1937984910
83.5333097210
91.1951002710
...
psi2
_112 cos _k( )( )4 sin 2_k( )( )2
0
4.4478 108
psi2
3 sin 2_k( )( )2_11 _21( )
0
25.4908 10
7 psi
2
sin _k( )( )4
sin 2_k( )( )2
_21( )2
0
2.7449 107
psi
Note -unable to directly use _avg1 max _11 _21 _11 _21( ) due to inner product being taken.require element by element operation. To combine all three vectors and filter for maximum value as shownbelow:
augment _11 _21 _11 _21( )
0 1 2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
42.050810 41.525910 35.248710
42.903910 42.160710 37.432210
42.592810 41.929210 36.635910
43.629910 42.700810 39.290310
42.178410 41.620910 35.575410
43.695410 42.749610 39.458110
42.227910 41.657710 35.70210
42.280210 41.696610 35.835810
42.899410 42.157410 37.420810
42.055510 41.529410 35.260910
41.915210 41.42510 34.901610
42.21610 41.648910 35.671710
43.649110 42.715110 39.339410
42.526310 41.879710 36.465810
43.45710 42.572210 38.847910
42.880910 42.143510 ...
psi
7/30/2019 Mathcad - Example 8
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Combine three vectors into matrix for _11 _21 _11 _21 and choose maximum value and output as vector
i 0 rows augment _11 _21 _11 _21( )( ) 1
j 0 cols augment _11 _21 _11 _21( )( ) 1
Max_vali
max submatrix augment _11 _21 _11 _21( ) i i 0 cols augment _11 _21 _11 _21( )( ) 1( )( )
Max_val
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
42.050810
42.903910
42.592810
43.629910
42.178410
43.695410
42.227910
42.280210
42.899410
42.055510
41.915210
42.21610
43.649110
42.526310
43.45710
...
psi
Calculate _1
_1 _avg1 Max_val( )
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
41.728510
...
psi
7/30/2019 Mathcad - Example 8
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Calculate E_avg1
Note: Currently no filter to limit value to 1
Check
E_avg_1_1
_1( )0.5
0
0
1
2
3
4
5
6
7
8
9
10
11
1213
14
15
0.8461
0.6006
0.68
0.4951
0.7942
0.4719
0.7885
0.8341
0.7115
0.8409
0.906
0.7928
0.50030.9196
0.5
...
_1
_1( )0.5
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
0.8461
0.6006
0.68
0.4951
0.7942
0.4719
0.7885
0.8341
0.7115
0.8409
0.906
0.7928
0.5003
0.9196
0.5
0.612
0.6382
0.7458
0.672
Calculate RSF_1
Check W_avg_10
0.45 in
T_c 1.045 in
E_avg_10
0.8461
Note that 0.9595 value is different from example
value,0.7255 (API 579 2007 edition)RSF_1 1
W_avg_1 1 E_avg_1( )[ ]
T_c
0.9337
0.7611
0.8086
0.6859
0.8917
0.76
0.8735
0.9087
0.9034
0.901
0.9595
0.8562
0.7011
0.9711
0.7129
0.7308
0.8788
0.8541
0.8274
7/30/2019 Mathcad - Example 8
11/20
Calculate RSF for region of pittingCheck
length RSF_1( ) 19
RSF_pit
1
i
RSF_1i
length RSF_1( )
0.8379
i
RSF_1i
15.9197
Note that PSF_pit value is different from example due to one entry being different
RSF_pit 0.8379
RSF_a 0.9
Check longitudinal stress condition
Statement 3
result_3 if RSF_pit RSF_a true false( )
result_3 "true"
Hence determine MAWP_r using equation in Part2 paragraph 2.4.2.2
MAWP_r MAWPRSF_pit
RSF_a
471.2764 psi
7/30/2019 Mathcad - Example 8
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Calculate R_t
R_tT_c FCA w_ik
T_c
0
0
12
3
4
5
6
7
8
9
10
11
12
13
14
15
0.5694
0.47370.5694
0.378
0.4737
0.5694
0.4258
0.6651
0.5694
0.5215
0.6651
0.3301
0.5694
0.8086
0.378
...
Calculate Q
From table 4.5
Q 1.1231 R_t( )
1 R_tRSF_a
2
1
0.5
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.6869
0.5439
0.6869
0.435
0.5439
0.6869
0.4865
0.9028
0.6869
0.6095
0.9028
0.3878
0.6869
1.7942
0.435
...
Check
R_t0
0.5694
RSF_a 0.9
7/30/2019 Mathcad - Example 8
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Determine if
Check
D 60 in
T_c 1.045 in
Q0
0.6869
Q D T_c( )0.5
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
5.4388
4.3067
5.4388
3.4447
4.3067
5.4388
3.8523
7.1489
5.4388
4.8264
7.1489
3.0704
5.4388
14.2069
3.4447
...
in d values d_ik
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.5
1.6
0.9
1
0.7
1.1
0.8
0.5
1.3
0.4
1.5
0.6
2.4
0.4
1
...
in
Output 1 if the above statement is true
d_ik Q D T_c( )0.5
0
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Manually check through to determine!
Output false if there is a min of 0
All pits passed?
if min d_ik Q D T_c( )0.5
0= false true "true"
7/30/2019 Mathcad - Example 8
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All values calculated before
D 60 in
D_o 62.25 in
FCA 0.05 in
LOSS 0.03in
T_rd 1.095 inR_c 30.08 in
T_c 1.045 in calculated from before
RSF_pit 0.8379
MAWP_r 471.2764 psi
RSF_a 0.9
T_sl 0.3633in
Weight Case
Q_s 3.469 104
lbf
M 1.3126 106
in lbf
No Thermal Load assumed
7/30/2019 Mathcad - Example 8
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Calculate B
B min 1 RSF_pit
RSF_a
0.931
Calculate T_eq
T_eq B T_c 0.9729 in
No thermal load case.
Weight Case
Q_s 3.469 104
lbf
M 1.3126 106
in lbf
From 2007 API example manual
_cm
MAWP_rR_c
T_eq0.6
RSF_pit cos ( )1.7728 10
4 psi
7/30/2019 Mathcad - Example 8
16/20
CheckD_f D_o 2T_eq 60.3043 in
D_o 62.25 in
T_eq 0.9729inA_m
4
D_o2
D_f2
187.2856 in2D_f 60.3043 in
D_o 62.25 in
aD_o
231.125 in
I_x
64
D_o4
D_f4
8.7927 104 in4
A_t
16 D_o D_f ( )2 2.9491 103 in2
A_a
4
D_f2
in2
Calculate
Shear Stress- no torsion loading and shear load at midspan is zero
M_t
a A_t T_eq
V_l
A_m 0 psi Check
A_m 187.2856in2
T_eq 0.9729in
A_t 2.9491 10
3
in2
Longitudinal Membrane Stress
No applied section axial force for weight or weight plus thermal load
7/30/2019 Mathcad - Example 8
17/20
Calculate
Tensile
_lmtA_a
A_m
MAWP_rF_axial
A_m
M a
I_x
1
E_c cos ( )
9.0021 103
psi
Check A_m 1.3006 ft2
A_a 2.8562 103
in2
A_m 187.2856in2
F_axial 0 lbf
A_m 187.2856in2
M 1.3126 106
lbf in
I_x 8.7927 104
in4
a 31.125 in
Compressive
_lmcA_a
A_m
MAWP_rF_axial
A_m
M a
I_x
1
E_c cos ( )
7.9088 103
psi
Weight case-tensile
_et _cm2
_cm _lmt _lmt2
32
0.5
1.5354 104
psi
Check 0
_lmt 9.0021 103
psi
_cm 1.7728 104
psi
Weight case-compressive
_ec _cm2
_cm _lmc _lmc2
32
0.5
1.5383 104
psi
No Thermal Loads
7/30/2019 Mathcad - Example 8
18/20
H_f assumed to be 1 for weight case
H_f 1 Check
SA 1.75 104
psiH_f
SA
RSF_a
1.9444 104
psiRSF_a 0.9
max _et _ec( ) 1.5383 104
psi_et 1.5354 10
4 psi
_ec 1.5383 104
psi
Statement 4
result_4 if max _et _ec( ) H_f
SA
RSF_a
true false
result_4 "true"
Conclusion from API 579 2007
If maximum longitudinal stress in step 7 is compressive, this stress should be less than or equal to allowable
compressive stress computed using paragrpah A.4.4 or allowable tensile stress, whichever smaller.When using
this methodology to establish an allowable compressive stress, an average thickness representative of the region
of pitting damage in the compressive stress zone should be used in the calculations.
The maximum longitudinal stress in STEP 7 is NOT compressive.
SUMMARY
MAWP_r 471.2764 psi
The longitudinal stress is acceptable. Equipment fir for service for MAWP_r calculated.
7/30/2019 Mathcad - Example 8
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