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Structural analysis and design of pipelines
different loadings
internal pressure (operating pressure)
external pressure (hydrostatic pressure of sea)
temperature changes (operating)
bending (construction, spans, upheaval)
concentrated loads (construction, accidents)
impact (accidents)
0.75 m
15 MPa
0.75 m
15 MPa
wall tension
1 m long
=
= (pressure)(area)
= (15 MN/m2)(0.75 m2 )
= 11.25 MN
= 1100 tonnes
vertical equilibrium of half pipe and half contentsresultant force must be zero
D
p
po
i
i
Do
t t
D
p
po
i
i
Do
t t
D
ps
i
Do
t t
H
i
po
sH
iiHoo DptsDp =+ 2
vertical equilibrium of unit length
p D s t p D
sp D p D
ts
o o H i i
Hi i o o
H
+ =
=−
2
2
rearrange
exact for mean value of
D/tt
tDpps
t
Dps
st
DpDps
ooiH
iH
H
ooiiH
of valueslfor typica small are sdifference
1996) (DnV 2
))((
(Barlow) 2
variants
of mean valuefor exact 2
−−=
=
−=
design
most codes require that the hoop stress is less than a prescribed fraction f1 of the yield stress Y
s f YH ≤ 1
stressUY
0.005 strain
f1Y
f1 is called the design factor (usage factor)
historically f1 was taken as 0.72 for pipelines, 0.60 or less for risers
those factors date back 70 years, to a time when standards of design, pipe manufacture, welding and construction were far inferior to today
recent thinking is that higher factors to and above 0.8 can be applied (provided that the governing code allows that change)
rearrange
Do is 762 mm (30 inches), pi is 20 MPa (200 bars, 2900 psi) po is 2 MPa (20 bars, 290.1 psi) Y is 413.7 MPa (N/mm2) (60000 psi,
X60), f1 is 0.83, t > 18.87 mm (0.743 inches)
)(2
)(2
)2(
2
1
00
01
i
i
ooiooiiH
pYf
Dppt
t
DptDp
t
DpDpsYf
+−≥
−−=−=≥
example
12 1 +
−
≥
oi pp
YfD
t
DnV 1996 formula
rearrange
D is 762 mm (30 inches), pi is 20 MPa (200 bars, 2900 psi) po is 2 MPa (20 bars, 290.1 psi) Y is 413.7 MPa (N/mm2) (60000 psi,
X60), f1 is 0.83, t > 19.46 mm (0.766 inches)
t
tDppsYf oi
H 2
))((1
−−=≥
s p pD D
D DpH i o
o i
o io= − +
−−( )
2 2
2 2
exact for linear elastic material, even for thick-walled tube
Lamé formula
allowed by some codes
Lamé formula
design
t Do= − −
12
1 14β
β = + +−
2 1 1f Y p
p po
i o
where
with the same values as in the previous example
t = 19.37 mm (0.763 inches)(compare with 18.87 mm from exact equation for mean stress, and 19.46 mm from DnV 1996)
p pp t
li eb
SC m
− ≤ ( )1
γ γ
γSC is a safety class reduction factor, listed in table 5-5 of the rules γm is a material resistance factor, listed in table 5-4
DnV 2000 and 2007 rules adopt a limit state approach
allowable pressure difference (inside – outside) is the burst pressure divided by two factors
where pb is the burst pressure
−−=
3
2
15.1
2,
3
22Min
U
tD
tY
tD
tpb
the burst pressure is
where
Y is the characteristic yield stress (fy in DNV 2007)
U is the characteristic ultimate tensile strength(ft in DNV 2007); Y governs if U > 1.15Y
stressUY
0.005 strain
−+
=
15.1,Min
3
2
)(
21
UY
pp
Dt
elimSCγγ
continuing the example, and taking
γsc 1.138 (safety class normal)γm 1.15Y 413.7 MPa (60000 psi)U >1.15Y (so that Y governs)
minimum wall thickness is 18.34 mm (0.722 inch)
(rearrange)
9092949698
100102104106
400 450 500 550 600
SMYS (MPa)
% c
ost (
refe
rred
to g
rade
48
3 ba
se c
ase)
120 bars
80 bars
high-strength steels
TransCanada (TCPL) started using grade 483 (X70) 30 years ago, now has 6300 kmstarted using grade 550 (X80) in 1991, now has 400 kmnow introducing grade 690 (X100), 1 km installed in 2002
project cost reduction by higher grades and higher operating pressure
external pressure
complex interaction between elastic circumferential bending, plastic circumferential bending and initial out-of-roundness
DnV 2007 clause 5D 401
ptRgpppppp YecrYecr )/(2))(( 22 =−−
pecr = 2E/(D/t)3(1-ν2) is the pressure at which the pipe would collapse if it remained elastic and yield did not occur
pY = 2Yαfab /(D/t) is the pressure at which the hoop stress would reach αfabY in compression if the pipe remained round and elastic instability did not occur
g is the out of roundness (max-min)/mean diameter, and >0.005αfab is a fabrication factor, 1 for seamless, 0.85 for UOE
p is the collapse pressure (and must be smaller than the smaller of pecr
and pY)
The DNV 2007 notation is slightly different
α β αβ α α3 5 3 2 1 0( / ) ( / ) ( )( / )D t D t g D t− − + + =
where
α
β ν
=
= −
p
Y
Y
E
2
1 2( )
which is a quintic in D/t
0
5
10
15
20
25
30
35
0 0.01 0.02 0.03 0.04 0.05
p/2Y
D/t
g=0.01
g=0.02
g=0.03
0 0 0 0 0 0 0
5
3
10
1 1
0123456
0 to5
5 to10
10to15
15to20
20to25
25to30
30to35
35to40
40to45
45to50
50to55
55to60
60to65
collapse pressure (MPa)
num
ber
of te
sts
calculated
alternatively
API RP1111
pp p
p p
ecr Y
ecr Y
=+2 2
which does not include out-of-roundness, but for reasonable values of g gives about the same collapse pressure as DnV 2000
sL
sH
hoop stress sH is statically determinate
longitudinal stress sL is not statically determinate, and depends on whether the pipeline moves longitudinally
quantify
idealise pipeline as thin-walled tube, radius R, wall thicknesst
elastic modulus E
Poisson’s ratio ν
thermal expansion coefficientα
operating temperature rise θ (operating - installation)
operating pressure p
Rt
material elastic, isotropic;
stress and strain tensile positive
general relationship for longitudinal strain
εν
αθLL Hs
E
s
E= − +
longitudinal hoop thermal
θαν
ε
Et
pRs
tpRtpDs
L
H
L
−=
===
then
on)idealisati (thin wall /2/and
0if
εν
αθLL Hs
E
s
E= − +
longitudinal hoop thermal
s longitudinalL
hoopsH
Y
Y
von Mises yield condition for biaxial combined stress
s longitudinalL
hoopsH
Y
Y
von Mises yield condition for biaxial combined stress
pressure
temperature
s longitudinalL
hoopsH
Y
Y
von Mises yield condition for biaxial combined stress
pressure +temperature
L
H
some older codes impose a second limit on a von Misesequivalent stress that includes both sH and sL
(sH2 - sHsL + sL
2)1/2 = seqvM < f3Y
s
s
0
5
10
15
20
25
0 50 100 150
temperature rise (deg C)
req
uir
ed w
all
thic
knes
s (m
m)
equivalent stresscondition
hoop stresscondition
example
OD 323.85 mm (nominal 12 inch), operating pressure 20 MPa (200 bars, 2900 psi), X60, no temperature derating of yield stress
the equivalent stress condition governs the design of constrained high-temperature pipelines, and leads to a big increase in wall thickness
recent thinking is that this condition is not necessary
some plastic strain occurs, but it is limited in magnitude and does not lead to instability or rupture
wall thickness can then be reduced
allowable-strain design permitted by ANSI B31.4, DnV 1996, DnV 2000, DnV 2007, BS 8010 Part 3, NEN3650, CSA Z662
safeguards:
limit magnitude of plastic strain
do not apply to large D/t (< 60 for BS8010, <45 for DnV 2007)
make sure welds have adequate ductility
take account of reduced flexural rigidity
curvature
bending moment
first yield
buckling
curvature localises
1
2
bending buckling
0.000
0.010
0.020
0.030
0.040
0 10 20 30 40 50 60
D/t
bu
cklin
g s
trai
n
tests
DNV (no pressure, alphafactors 1)
comparison between data and DNV formula (DNV 2007 5D 608) for buckling strain
external pressure reducesthe critical curvature at which pipe buckles
internal pressure increasesthe critical curvature at which pipe buckles
if bending and large external pressure occur together, another phenomenon becomes important
2/5
6
=R
tYpprop
propagation is initiated by a combination of bending and pressure
once started, the buckle can propagate at a lower pressure
propagation pressure is minimum pressure at which a buckle can continue to propagate
proportional to t5/2
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0 10 20 30 40 50
D/t
p/Y data
DNV formula (afab = 1)
comparison between data and DNV formula (DNV 2007 5D 501) for propagation pressure
alternative strategies to deal with buckle propagation
one
increase wall thickness so that
propagation pressure > maximum external pressure
two
accept possibility of propagation over short distances, but incorporate buckle arresters, so that buckle runs to arresters on either side and then stops
Sleeve arresterIntegral arrester
Ring arrester
Wierzbicki analysis
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 10 20 30 40 50
deflection (mm)
inde
ntat
ion
forc
e (M
N)
theory
measured812.8 mm (32-inch), 19.05 mm wall, X65
impact
need to know when a missile will rupture a pipe
tests by Neilsen (AEA Winfrith)
U
D
t
d
for flat-ended missiles, the missile kinetic energy at which the pipe is just perforated is
3
50751751
kJ/mm 00560.cY
DdcYtE ...
=
= −
for schedule 40
pointed missiles perforate at lower energiesliquid-filled pipes perforate at lower energies
0
20
40
60
80
100
0 50 100 150 200
impact energy (kJ)
redu
ctio
n in
ID (
mm
)
on sand
on steel
1 23
8
7
6
54