s 12
SIF Solutions
for some more PRACTICAL configurations involving Chapter
pipers
and
634
cylinders
12.3.1
Stress
Intensity
Factor
TABLE 12.29
Stress intensity solutions for circumferential through-wall flaws in cylinders [7].
Axial Tension.
= at..J n:RB Ft
KJ
p
at= 2nRt
where
1.5
B
= 1+ A[ 5.3303 ( n:)
Ft
A
= ( 0.125(-0.25
)
]
R
for 5::; -::; 10
t
0.25
R
A = ( 0.4(-3.0
Bending Moment.
)
= ab..Jn:RBFb
KJ
M
where
ab
= n:R2t
e
Fb
+ 18.773( n:)
0.25
R
4.24
B
= 1+ A [ 4.5967
()
1.5
n
e
+ 2.6422( n
4.24
)
]
where A is as defined above for the pure tension case.
Internal Pressure.
KJ
= am..Jn:RBFm
R - mean radius:
where
R
= Ro + Ri
2
Fm
am
=
pR
2t
= 1+0.1501y1.5
for
y::;2
e -crack half angle in radians
Fm =0.8875+0.2625y
y=B~
for 2 ::;y ::;5
SIF for Pipes and Cylinders (Continued)
635
K, J, Compliance, and Limit Load Solutions
TABLE 12.30
Stress intensity solutions for part-through internal circumferential flaws in cylinders subject to uniform
tension [11]
Axisymmetric
Flaw.
= (Jt-Y 1T:aFt
KJ
a 1.5
= 1.1+ A [ 1.948
Ft
(t )
~3342(
R
A
r]
0.25
= ( 0.125-t - O.25)
R
A
~
R
for 5 ~ -t ~ 10
0.25
= ( 0.4-f - 3.0)
R
for 10 ~
-f ~ 20
Semielliptical Flaw. (Kf at deepest point.)
~
KJ
= (Jt~QFt
Ft
= 1+ [0.02
+ ~(0.01O3 + 0.00617~)
~.OO3S(I+O.7g)(~
-sr7]Q2
~=~t
(
!!:.-
2c
1.65
)
()
Q=1+1.464 ~
--..
SIF for Pipes and Cylinders (Continued)
636
Chapter 12
Stress intensi
= ah~1ta
K[
J1;0.52X+1.29x2
ah
-0.074X3
= pR
t
a
X = -:;;Rt
Long Part-Through Internal Raw.
5 ~ R/t ~ 20 and alt ~ 0.750.
2
2pR
K[
= R2 0
0
a R-
r=
2 "V1taF
R.l
(
-,-1..
t
F=1.l++.95tr
A
Valid for
t
)
+1.092(;r]
=( 0.125~i - 0.25t25 for 5 ,; ~i ,; 10
0.25
R.
A= ( 0.2-;--1.0
)
Finite Length Part-Through Internal Flaw. Valid
for 5 ~ R/t ~ 20, 2c/a ~ 12, and aft ~ 0.80. (Kf at
deepest point.)
(
Kj
t ~ n:a
Q F
= pR
F
= 1.12 + O.053~
~
!!:.,!!:.-,R
t 2c t )
+ O.0055~2
+ (1+ 0.02~ + 0.0191~2)
~ = at
(
20--
t
1400
!!:.-
( 2c )
Q
= 1 + 1.464(;
R 2
)1.65
)