Chapter 6 Application of Developed Formulations
6.2 Effects of pipe restraint on LBB evaluation
6.2.2 Evaluation methods of the pipe restraint effects on LBB150
To investigate the effect of the pipe restraint on LBB design, the PEDs were derived for the case of i) using current LBB procedure (restraint is not considered), ii) considering only the restrained COD and iii) considering both the restrained COD and effective applied moment, respectively. The detailed descriptions of the calculation method are following.
i) Crack opening displacement
The COD of an unrestrained pipe was calculated using the formula in the ductile fracture handbook (Zahoor, 1989). The restrained COD was determined based on the calculation process in the subsection 5.1.1.1 using the linear elastic restraint coefficient.
ii) Leakage size crack
The leak rate for a postulated crack length and COD was calculated using the PICEP code developed by Electric Power Research Institute (Norris et al., 1984). When the calculated leak rate equals to 10 gpm, the leakage size crack was determined.
151
iii) Crack stability analysis
Because the material of the model was austenitic stainless steel base metal, the limit load formula in Eq. 6.1 was used for the crack stability analysis according to the Standard Review Plan 3.6.3 (US NRC, 2007b).
4 4
,
2
2 2 sin sin
4
0.5 /
2
o i f
instability unrestrained
m
m f
m
m in
m
R R
M R
where P
P P R R t
(6.1)
Minstability,unrestrained, Pm, Pin and σf denotes the instability moment for unrestrained pipe, the membrane stress, the internal pressure, and the flow stress of the material, respectively. To consider the restraint effect on the crack stability analysis, Eq. 6.1 was corrected as:
4 4
,
, ,1 ,
2 , ,1 ,
1 2
2 sin sin 4
0.5 /
2
o i f
instability restrained
Rest M D LE m
m f
m
m Rest M D LE in
m
R R
M C R
where P
P C P R
R t
(6.2)
where the increase of the load-carrying capacity of a crack was reflected to Pin and Minstability,restrained.
The geometries and material of pipe and the operating conditions were referred from the typical reactor coolant system of PWR. Other information about the evaluation matrix and material properties are summarized in Table
6.2.
6.2.3 Evaluation results
Figure 6.9 shows the leakage size crack and the piping evaluation diagram of each case depending upon the normal operating moment (MNOP). The instability moment and the normal operating moment were normalized by a limit of the design moment (MASME). The limit of design moment was defined using Equation 10 of NB-3600, ASME B&PV Code Section III (ASME, 2010b) which is the requirement for piping systems under the normal operating conditions.
1 1 2
2
2 3 , 1
2
in O
ASME m
O
P D
M I S C C C
C D t
(6.3)
Sm and Pin are the design stress intensity and the internal pressure, respectively. C1 and C2 are secondary stress indices. MASME was derived assuming that the stress term induced by thermal gradient is zero.
Graphs (a) to (f) of Figure 6.9 show the results for the 12 inch pipe. If the pipe restraint effect on COD is considered, the length of leakage size crack increases due to the narrowed flow path. When only the restrained COD was considered (blue lines), instability loads were lower than those predicted through the current LBB method (black lines, not considering pipe restraint effect) at the same MNOP. However, if the restrained COD and the effective applied moment at the cracked section were considered simultaneously (red
153
From this, it was confirmed that the pipe restraint has great influence mainly by the increase of the load-carrying capacity, rather than the reduction of the COD.
The degree of the restraint effect increased as the crack length increased or restraint length decreased. This corresponds to decreases in the value of the restraint coefficient. Regarding the restraint length, the restraint effect in (5,5) was insignificant than (1,20) or (1,20). This indicates that the restraint effect is dominated by the restraint length of shorter side. These trends are observed for both pipe diameter cases.
The objective of the example LBB analysis is to evaluate the combined results of effects of restrained COD and increase of the load- carrying capacity. The analysis results indicated that the restraint effect on the applied moment has more significant influence on the LBB evaluation than the restraint effect on COD. Therefore, the current LBB evaluation procedure, with no attention to the pipe restraint effect, can predict conservative results compared to the case in which the restraint effect is considered for the conditions examined herein. In addition, if the restraint effect is implemented into the current practice of deterministic LBB analysis using the developed formulations, the piping system can be shown to possess greater safety margins. Because the value of linear elastic restraint coefficient is greater than the elastic-plastic restraint coefficient, when plastic deformations occur, the margin might actually be more significant.
Table 6.1 Matrix of analysis for calculation of COD and J-integral
Type Rm [in] Rm/t Internal Pressure [psi] Crack length [θ/π] Restraint length (L1/Do, L2/Do) Symmetric
5.72 5 0, 2320 0.125, 0.25, 0.5
(5,5), (10,10), (20,20)
Asymmetric (1,10), (1,20)
155
Table 6.2 Matrix of analysis and material properties used for LBB evaluations
Type
Pin
[psi]
Temp.
[°F]
Rm
[in]
Rm/t
Restraint length (L1/Do , L2/Do)
Material E [psi]
α n
σy
[psi]
σu
[psi]
12inch Pipe
2320 563
5.72 5 (1,1), (5,5), (10,10) (1,5), (1,10), (1,20)
TP316 (327℃)
2.73∙107 2.361 7.848 23,150 67,380
16inch Pipe 7.2 5
Figure 6.1 Summary of analysis case to demonstrate the applicability of the restraint coefficient in LBB design
Not considering
pipe restraint effect Case 1
Considering pipe restraint effect
Case 2
Case 3
157
Figure 6.2 3D FE model of pipe with a circumferential through-wall crack used for COD and J-integral calculations
Figure 6.3 Tensile property of TP316 stainless steel
L
1L
2Cracked Section x
z y
0.00 0.02 0.04 0.06 0.08 0.10
0 1x104 2x104 3x104 4x104 5x104
True Stress [psi]
True Strain
TP 316 SS 327 C
(a) Symmetric model – θ/π=0.125
(b) Symmetric model – θ/π=0.25
Figure 6.4 Comparisons of COD to validate the restraint coefficient
0.0 4.0x105 8.0x105 1.2x106 1.6x106
0.000 0.002 0.004 0.006 0.008 0.010
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=0 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
0.0 3.0x105 6.0x105 9.0x105 1.2x106
0.000 0.005 0.010 0.015 0.020
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin=0 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
159
(c) Symmetric model – θ/π=0.5
(d) Asymmetric model – θ/π=0.125
Figure 6.4 Comparisons of COD to validate the restraint coefficient (Internal pressure was not included) (Continued)
0.0 2.0x105 4.0x105 6.0x105 8.0x105
0.00 0.01 0.02 0.03 0.04 0.05 0.06
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin=0psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
4.0x105 8.0x105 1.2x106 1.6x106
0.000 0.002 0.004 0.006 0.008 0.010
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=0
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
(e) Asymmetric model – θ/π=0.25
(f) Asymmetric model – θ/π=0.5
Figure 6.4 Comparisons of COD to validate the restraint coefficient
0.0 3.0x105 6.0x105 9.0x105 1.2x106
0.000 0.003 0.006 0.009 0.012 0.015
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin= 0 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
0.0 2.0x105 4.0x105 6.0x105 8.0x105
0.000 0.005 0.010 0.015 0.020 0.025 0.030
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin= 0 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
161
(a) Symmetric model – θ/π=0.125
(b) Symmetric model – θ/π=0.25
Figure 6.5 Comparisons of J-integral to validate the restraint coefficient (Internal pressure was not included)
0.0 5.0x105 1.0x106 1.5x106 2.0x106 2.5x106
0 200 400 600 800 1000
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=0 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
0.0 5.0x105 1.0x106 1.5x106 2.0x106
0 300 600 900 1200 1500
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin=0 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
(c) Symmetric model – θ/π=0.5
(d) Asymmetric model – θ/π=0.125
Figure 6.5 Comparisons of J-integral to validate the restraint coefficient
0.0 3.0x105 6.0x105 9.0x105 1.2x106 1.5x106
0 300 600 900 1200 1500
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin=0psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
0.0 5.0x105 1.0x106 1.5x106 2.0x106 2.5x106 3.0x106
0 200 400 600 800 1000
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=0 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
163
(e) Asymmetric model – θ/π=0.25
(f) Asymmetric model – θ/π=0.5
Figure 6.5 Comparisons of J-integral to validate the restraint coefficient (Internal pressure was not included) (Continued)
0.0 5.0x105 1.0x106 1.5x106 2.0x106 2.5x106
0 200 400 600 800 1000 1200
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin= 0 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
0.0 3.0x105 6.0x105 9.0x105 1.2x106 1.5x106
0 200 400 600 800 1000 1200
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin= 0 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
(a) Symmetric model – θ/π=0.125
(b) Symmetric model – θ/π=0.25
Figure 6.6 Comparisons of COD to validate the restraint coefficient
0.0 3.0x105 6.0x105 9.0x105 1.2x106
0.000 0.002 0.004 0.006 0.008 0.010
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=2320 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
0.0 2.0x105 4.0x105 6.0x105 8.0x105 1.0x106
0.00 0.01 0.02 0.03 0.04
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin=2320 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
165
(c) Symmetric model – θ/π=0.5
(d) Asymmetric model – θ/π=0.125
Figure 6.6 Comparisons of COD to validate the restraint coefficient (Internal pressure was included) (Continued)
0 1x105 2x105 3x105 4x105 5x105 6x105
0.00 0.05 0.10 0.15 0.20
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin=2320 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
0.0 3.0x105 6.0x105 9.0x105 1.2x106
0.000 0.002 0.004 0.006 0.008 0.010
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=2320 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
(e) Asymmetric model – θ/π=0.25
(f) Asymmetric model – θ/π=0.5
Figure 6.6 Comparisons of COD to validate the restraint coefficient
0.0 2.0x105 4.0x105 6.0x105 8.0x105 1.0x106
0.000 0.005 0.010 0.015 0.020 0.025 0.030
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin=2320 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
0 1x105 2x105 3x105 4x105 5x105 6x105
0.00 0.05 0.10 0.15 0.20
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin=2320 psi
Crack opening displacement [in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
167
(a) Symmetric model – θ/π=0.125
(b) Symmetric model – θ/π=0.25
Figure 6.7 Comparisons of J-integral to validate the restraint coefficient (Internal pressure was included)
0.0 5.0x105 1.0x106 1.5x106 2.0x106 2.5x106
0 300 600 900 1200 1500 1800
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=2320 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
0.0 3.0x105 6.0x105 9.0x105 1.2x106 1.5x106
0 400 800 1200 1600
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin=2320 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
(c) Symmetric model – θ/π=0.5
(d) Asymmetric model – θ/π=0.125
Figure 6.7 Comparisons of J-integral to validate the restraint coefficient
0.0 3.0x105 6.0x105 9.0x105 1.2x106 1.5x106
0 2000 4000 6000 8000 10000
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin=2320 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[5:5] [10:10] [20:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
0.0 5.0x105 1.0x106 1.5x106 2.0x106 2.5x106
0 300 600 900 1200 1500
TP 316 base LB
= 0.125 Rm=5.72 inch Rm/t=5 Pin=2320 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
169
(e) Asymmetric model – θ/π=0.25
(f) Asymmetric model – θ/π=0.5
Figure 6.7 Comparisons of J-integral to validate the restraint coefficient (Internal pressure was included) (Continued)
0.0 5.0x105 1.0x106 1.5x106 2.0x106
0 300 600 900 1200 1500
TP 316 base LB
= 0.250 Rm=5.72 inch Rm/t=5 Pin=2320 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L
1/D
o:L
2/D
o) Case 1
Case 2 Case 3
0.0 3.0x105 6.0x105 9.0x105 1.2x106 1.5x106
0 3000 6000 9000 12000 15000
TP 316 base LB
= 0.500 Rm=5.72 inch Rm/t=5 Pin=2320 psi
J integral [psi-in]
Applied moment at the crack postion of uncracked pipe [lbf-in]
[1:10] [1:20] (=L1/Do:L2/Do) Case 1
Case 2 Case 3
Figure 6.8 Schematic diagram of the piping evaluation diagram
LBB Satisfied
Margin of 1.4 on MFault
Margin of 2.0 on θl
M
NOPM
Fault,Allow171
(a) 12 inch, L1/Do=1, L2/Do=5
(b) 12 inch, L1/Do=1, L2/Do=10
Figure 6.9 Effect of the restrained COD and the effective applied moment on LBB evaluation
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=12 inch, R/t = 5 L1=1D, L2=5D MASME=4.01E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=12 inch, R/t = 5 L1=1D, L2=10D MASME=4.01E+6 [lbf-in]
M Instability/M ASME
MNOP/MASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
(c) 12 inch, L1/Do=1, L2/Do=20
(d) 12 inch, L1/Do=1, L2/Do=1
Figure 6.9 Effect of the restrained COD and the effective applied moment on LBB evaluation (Continued)
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=12 inch, R/t = 5 L1=1D, L2=20D MASME=4.01E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=12 inch, R/t = 5 L1=1D, L2=1D MASME=4.01E+6 [lbf-in]
M Instability/M ASME
MNOP/MASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
173
(e) 12 inch, L1/Do=5, L2/Do=5
(f) 12 inch, L1/Do=10, L2/Do=10
Figure 6.9 Effect of the restrained COD and the effective applied moment on LBB evaluation (Continued)
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=12 inch, R/t = 5 L1=5D, L2=5D MASME=4.01E+6 [lbf-in]
M Instability/M ASME
MNOP/MASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=12 inch, R/t = 5 L1=10D, L2=10D MASME=4.01E+6 [lbf-in]
M Instability/M ASME
MNOP/MASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
(g) 16 inch, L1/Do=1, L2/Do=5
(h) 16 inch, L1/Do=1, L2/Do=10
Figure 6.9 Effect of the restrained COD and the effective applied moment
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=16 inch, R/t = 5 L1=1D, L2=5D MASME=7.99E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=16 inch, R/t = 5 L1=1D, L2=10D MASME=7.99E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
175
(i) 16 inch, L1/Do=1, L2/Do=20
(j) 16 inch, L1/Do=1, L2/Do=1
Figure 6.9 Effect of the restrained COD and the effective applied moment on LBB evaluation (Continued)
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=16 inch, R/t = 5 L1=1D, L2=20D MASME=7.99E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=16 inch, R/t = 5 L1=1D, L2=1D MASME=7.99E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
(k) 16 inch, L1/Do=5, L2/Do=5
(l) 16 inch, L1/Do=10, L2/Do=10
Figure 6.9 Effect of the restrained COD and the effective applied moment on LBB evaluation (Continued)
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=16 inch, R/t = 5 L1=5D, L2=5D MASME=7.99E+6 [lbf-in]
M Instability/M ASME
MNOP/M
ASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
0.0 0.2 0.4 0.6 0.8 1.0
0.00 0.25 0.50 0.75 1.00 1.25 1.50
TP 316
Press=16MPa, Temp.=295ºC D=16 inch, R/t = 5 L1=10D, L2=10D MASME=7.99E+6 [lbf-in]
M Instability/M ASME
MNOP/MASME
Using current LBB method Considering the restrained COD Considering the restrained COD & the effective applied moment
0.0 0.2 0.4 0.6
LSC 2*LSC
Using current LBB method Considering the restrained COD
Leakage Size Crack []
177