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Spherical pressure vessel 1
- Thread starter Hlengani
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Don't know if this helps; if you have access to NASGRO, you can find solutions for through cracks and surface cracks in spheres.
Tada-Paris also has one, for through crack length '2a' in a sphere: KI=sigma*sqrt(pi*a)*F(lambda)
where KI==Mode I Stress Intensity Factor
sigma==0.5*p*R/t (p=internal pressure, R=radius, t=
shell thickness)
F(lambda)=sqrt(1+1.41*lambda^2+0.04*lambda^3)
lambda==a/sqrt(R*t)
Restrictions? 0<lambda<=3
Hope you understand the clumsy way I have written this.
How is an axial crack defined in a sphere?
Tada-Paris also has one, for through crack length '2a' in a sphere: KI=sigma*sqrt(pi*a)*F(lambda)
where KI==Mode I Stress Intensity Factor
sigma==0.5*p*R/t (p=internal pressure, R=radius, t=
shell thickness)
F(lambda)=sqrt(1+1.41*lambda^2+0.04*lambda^3)
lambda==a/sqrt(R*t)
Restrictions? 0<lambda<=3
Hope you understand the clumsy way I have written this.
How is an axial crack defined in a sphere?
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- #4
Prost thanx very much, the most NB thing that i was looking for was the factor [F(lambda)] in other words the shape factor for through crack in a sphere. I am sorry for mentioning axial crack, i was confusing a longitudinal cylider with a sphere. cheers for now
then i think the relevant stress is the standard hoop stress (pR/t), but then maybe you know this.
also, for bulging effects in a fuselage i use rooke and cartwright, fig 197, for which my curve fit is ...
?fig197 = Gm+Gb
= [0.725+0.85*(a/(Rt)^0.5)] + [0.4-0.184*(abs(a/(Rt)^0.5-1.55))^1.77]
This curve fit is accurate for 0.5 < a/(Rt)^0.5 < 3.1.
For a/(Rt)^0.5 < 0.5, assume ?fig197 = 1.0
For a/(Rt)^0.5 > 3.1, ?fig197 = Gm-Gb
= [0.725+0.85*(a/(Rt)^0.5)] - [0.4-0.184*(abs(a/(Rt)^0.5-1.55))^1.77]
also, for bulging effects in a fuselage i use rooke and cartwright, fig 197, for which my curve fit is ...
?fig197 = Gm+Gb
= [0.725+0.85*(a/(Rt)^0.5)] + [0.4-0.184*(abs(a/(Rt)^0.5-1.55))^1.77]
This curve fit is accurate for 0.5 < a/(Rt)^0.5 < 3.1.
For a/(Rt)^0.5 < 0.5, assume ?fig197 = 1.0
For a/(Rt)^0.5 > 3.1, ?fig197 = Gm-Gb
= [0.725+0.85*(a/(Rt)^0.5)] - [0.4-0.184*(abs(a/(Rt)^0.5-1.55))^1.77]
I double checked Tada and Paris--the equations are in Tada-Paris's "Stress Analysis of Handbooks" prob. 36.1 as I have written. I think it matters not what the stress in a spherical pressure vessel actually is; the Tada definition of 'stress' can be anything he wants, as long as he defines it completely, right?
By my equilibrium calcs, the stress in the shell is pR/(2t), and this is the only stress. My mechanics book calls this the 'circumferential' or 'hoop' stress.
By my equilibrium calcs, the stress in the shell is pR/(2t), and this is the only stress. My mechanics book calls this the 'circumferential' or 'hoop' stress.
you're right, Tada-Paris can choose a stress definition that suits them ... but ...
even tho' the OP started talking about a spherical pressure vessel, he's changed that to a cyclindrical one. the solution i gave was for a longitudinal crack in a cyclindrical pressue vessel. Is there a different solution in Tada-Paris for this ?
even tho' the OP started talking about a spherical pressure vessel, he's changed that to a cyclindrical one. the solution i gave was for a longitudinal crack in a cyclindrical pressue vessel. Is there a different solution in Tada-Paris for this ?
rb1957: I believe the OP (original poster) made a mistake when he asked for axial crack solutions (see 4th post down from top), he really wanted just the one. Yes, Tada-Paris has some Equations for cylinders with cracks. These are all for pressurized cylinders (there are pipes im bending solutions if you need them):
For a single axial through crack (that is, crack oriented so it propagates in the axial direction):
a=half length of crack
sigma=p*R/t
lambda=a/sqrt(R*t)
KI=sigma*sqrt(pi*a)*F(lambda)
F(lambda)=sqrt(1+1.25*lambda^2) if 0<lambda<=1
or F(lambda)=0.6+0.9*lambda if 1<=lambda<=5
For a single circumferential through crack:
a=half length of crack
sigma=0.5*p*R/t 'Note difference in sigma definition!
lambda=a/sqrt(R*t)
KI=sigma*sqrt(pi*a)*F(lambda)
F(lambda)=sqrt(1+0.3225*lambda^2) if 0<lambda<=1
or F(lambda)=0.9+0.25*lambda if 1<=lambda<=5
Note: I have validated quite a few FE solutions against these Tada-Paris solutions for comparisons, but NOT for these particular sphere and cylinder solutions, so I cannot tell you how accurate these are. Tada Paris claims 1% on the KI
For a single axial through crack (that is, crack oriented so it propagates in the axial direction):
a=half length of crack
sigma=p*R/t
lambda=a/sqrt(R*t)
KI=sigma*sqrt(pi*a)*F(lambda)
F(lambda)=sqrt(1+1.25*lambda^2) if 0<lambda<=1
or F(lambda)=0.6+0.9*lambda if 1<=lambda<=5
For a single circumferential through crack:
a=half length of crack
sigma=0.5*p*R/t 'Note difference in sigma definition!
lambda=a/sqrt(R*t)
KI=sigma*sqrt(pi*a)*F(lambda)
F(lambda)=sqrt(1+0.3225*lambda^2) if 0<lambda<=1
or F(lambda)=0.9+0.25*lambda if 1<=lambda<=5
Note: I have validated quite a few FE solutions against these Tada-Paris solutions for comparisons, but NOT for these particular sphere and cylinder solutions, so I cannot tell you how accurate these are. Tada Paris claims 1% on the KI
Continuing with cracks in cylinders, just found this one, a NASA/FLAGRO equation.
Crack is 2c (surface) x a (into thickness), located on internal or external surface of a pressurized pipe. Crack grows axially, so that the hoop stress is the stress that opens it.
beta(a/c,a/t,theta)=0.97*(M1+M2*(a/t)^2+M3*(a/t)^4)*g*fphi*fc*fi
M1=1.13-0.09*a/c
M2=-0.54+0.89/(0.2+a/c)
M3=0.5-(1/(0.65+a/c))+14*(1-a/c)^24
g=1+(0.1+0.35*(a/t)^2)*(1-sin(theta))^2
fphi=((a/c)^2*cos(theta)^2+sin(theta)^2)^(1/4)
((cos(theta)^2 is cos(theta)*cos(theta))
theta here I think is still the elliptical angle
typically called phi, not theta.
fi=1 (internal crack) fi=1.1 (external crack)
fc=((1+k^2)/(1-k^2)+1-0.5*sqrt(a/t))*(t/(D/2-t))
where k=1-2t/D
t==thickness of cylinder, D=outside diameter of cylinder.
sigma=p*(D-2t)/2t
K1=beta*sigma*sqrt(pi*a)/PHI, PHI is equal to:
if (a/c<=1), PHI=1+1.464*(a/c)^1.65
if (a/c>1) PHI=1.464*(c/a)^1.65
(Love those functions, M3, with the 24th power!)
Crack is 2c (surface) x a (into thickness), located on internal or external surface of a pressurized pipe. Crack grows axially, so that the hoop stress is the stress that opens it.
beta(a/c,a/t,theta)=0.97*(M1+M2*(a/t)^2+M3*(a/t)^4)*g*fphi*fc*fi
M1=1.13-0.09*a/c
M2=-0.54+0.89/(0.2+a/c)
M3=0.5-(1/(0.65+a/c))+14*(1-a/c)^24
g=1+(0.1+0.35*(a/t)^2)*(1-sin(theta))^2
fphi=((a/c)^2*cos(theta)^2+sin(theta)^2)^(1/4)
((cos(theta)^2 is cos(theta)*cos(theta))
theta here I think is still the elliptical angle
typically called phi, not theta.
fi=1 (internal crack) fi=1.1 (external crack)
fc=((1+k^2)/(1-k^2)+1-0.5*sqrt(a/t))*(t/(D/2-t))
where k=1-2t/D
t==thickness of cylinder, D=outside diameter of cylinder.
sigma=p*(D-2t)/2t
K1=beta*sigma*sqrt(pi*a)/PHI, PHI is equal to:
if (a/c<=1), PHI=1+1.464*(a/c)^1.65
if (a/c>1) PHI=1.464*(c/a)^1.65
(Love those functions, M3, with the 24th power!)
- Thread starter
- #10
Hlengani:
Since R and t are the two parameters used to define the shell (be it spherical or cylindrical), what difference does it make? As far as I can tell, my version of Tada-Paris has only through crack solutions. And, almost all the stresses in the Stress Intensity Factor solutions are defined pR/t, which you may recall from mechanics of materials is the thin shell approximation for the hoop stress in a pressurized cylinder. Unfortunately, I do not know the history of these solutions, but if I had to guess, I would guess that these Tada Paris solutions are also thin shell approximations. which if true, should mean that any FEA computation you use to compare to the Tada Paris solution should be for a 'thin' shell.
If I had the time, I would just go get some of the references, such as Folias 1965 "An Axial Crack in a pressurized Cylindrical Shell," Int. Journal of Fracture Mechanics, Vol. 1, pp. 104-113, 1965
Since R and t are the two parameters used to define the shell (be it spherical or cylindrical), what difference does it make? As far as I can tell, my version of Tada-Paris has only through crack solutions. And, almost all the stresses in the Stress Intensity Factor solutions are defined pR/t, which you may recall from mechanics of materials is the thin shell approximation for the hoop stress in a pressurized cylinder. Unfortunately, I do not know the history of these solutions, but if I had to guess, I would guess that these Tada Paris solutions are also thin shell approximations. which if true, should mean that any FEA computation you use to compare to the Tada Paris solution should be for a 'thin' shell.
If I had the time, I would just go get some of the references, such as Folias 1965 "An Axial Crack in a pressurized Cylindrical Shell," Int. Journal of Fracture Mechanics, Vol. 1, pp. 104-113, 1965
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- #12
I have the following equation Y =(2/?)*[sec{(?/2)*(c/t)}], this correctional factot of Y is used for leak-before-break condition. so i would like to get the source of this equation or references. i am trying to establish if i can use this equation for sphere since it is said that is for plates.
not my area but, that correction looks a little funny to me ... c/t implies to me a part-thru thickness crack ... i'd have thought you try a thru-thickness crack (like prost's corrections are for) and show that the critical crack is detectable ...
With that "sec" (for secant) term in there, sure looks like a width correction of some sort. The equation doesn't print correctly on my screen. I see Y=(2/?)...and one
other question mark sec{(?/2)....I take it those are PI=3.1415...?
Admittedly, not an expert on LBB (leak before burst) concepts. However, I have a very nice book by Farahmand, chapter 3.7.5, that discusses LBB. You know the through crack solution, you know the part through crack solution, you know the material. Farahmand then says the LBB criteria are:
K(part through)=K(through)--that is, the part through crack grows until the 'a' (depth) crack length breaks through to the other side. Further criteria:
K(part through)<K1e
and
K(through)<Kc or K1c
K1e is the Part through fracture toughness (ASTM standard E740), Kc and K1c are the usual plane stress and plane strain fracture toughness. These last two conditions are stability conditions--is the K low enough that that crack growth is stable?
other question mark sec{(?/2)....I take it those are PI=3.1415...?
Admittedly, not an expert on LBB (leak before burst) concepts. However, I have a very nice book by Farahmand, chapter 3.7.5, that discusses LBB. You know the through crack solution, you know the part through crack solution, you know the material. Farahmand then says the LBB criteria are:
K(part through)=K(through)--that is, the part through crack grows until the 'a' (depth) crack length breaks through to the other side. Further criteria:
K(part through)<K1e
and
K(through)<Kc or K1c
K1e is the Part through fracture toughness (ASTM standard E740), Kc and K1c are the usual plane stress and plane strain fracture toughness. These last two conditions are stability conditions--is the K low enough that that crack growth is stable?
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Application of LEFM & FMDM Theory
Bahram Farahmand with George Bockrath and James Glassco
Chapters: 1-Intro to Fatigue and Fracture Mechanics
2-Conventional Fatigue (High- and Low-Cycle Fatigue)
3-LEFM
4-Fatigue Crack Growth
5-Fracture Control Program and NDI
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