Mathcad - Example 8

7/30/2019 Mathcad - Example 8

1/20

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.


2/20

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


3/20

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


4/20

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


5/20

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


6/20

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


7/20

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


8/20

_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


9/20

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


10/20

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


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


12/20

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


13/20

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"


14/20

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


15/20

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


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


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


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.


19/20


20/20

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