Understanding the Kellogg Equivalent Pressure Method for piping flanges
M
M
F
F
Remarks
1. This method is a gasket load calculation.
2. This paper counts the gasket load in terms of force per unit
length of circumference, as this approach is usual in the circular plate
theory.
3. For calculation of the reaction on free-to-rotate edge, the
circular plate theory gives results that are quite insensitive vs. variations
of plate rigidity.
a
G
G
The theory assumes that the flange is a circular plate.
F
bolts
H G0 =
HG0
HG0
a
∑ Fbolts
2π a
external loads.
is the gasket load for the case no pressure, no
G
No pressure, no external loads
F
HG (p)
H G ( p) = H G0 −
bolts
Fpressure
HG (p)
Fpressure
2π a
is the gasket load for the case internal pressure, no external loads.
Fpressure = π a 2 p
and
Internal pressure, no external loads
F
F
ap
2
Remark: A pressure load is decreasing the gasket reaction.
H G ( p) = H G 0 −
G=2a
bolts
H G ( F) = H G0 −
HG (F)
HG(F)
G=2a
F
2π a
is the gasket load for the case no pressure, external tensile force.
Remarks: - A tensile force F > 0 is decreasing the gasket load.
- A compressive force F < 0 is increasing the gasket load.
No pressure, external tensile force
-1-
Understanding the Kellogg Equivalent Pressure Method for piping flanges
M
F
maxHG (M)
A bending moment is changing the gasket load, but H G ( M ) is variable
along the gasket circumference.
The gasket load can be described by the following model, based on the
theory of circular plates.
bolts
minH G(M)
G=2a
M
No pressure, external bending moment
-p
p
a
G
V
-V
(Vmax)
(Vmin)
ds=adθ
dθ
θ
a cosθ
G=2a
pa
cos θ is the evaluation of the edge reaction
4
following the circular plate theory
(see “Theory of Plates and Shells” by S. Timoshenko and S.
V=
Woinowsky-Krieger, Second edition, 1959, paragraph 63
“Circular Plates under Linearly Varying Loads)
and
π
2
πa 3 p
 pa

M=4 
cos θ  (a cos θ) adθ =
4
4


0
∫
is the moment equation.
It results:
4M
p= 3
πa
pa 4 M a M
Vmax =
=
=
4 πa 3 4 πa 2
H G ( M ) = H G0 + Vmax cos θ = H G 0 +
M
cos θ
πa 2
M
θ
πa 2
is the minimum gasket load for the case no pressure,
external bending moment.
min H G ( M ) = H G0 −
-2-
Understanding the Kellogg Equivalent Pressure Method for piping flanges
THE FOLLOWING EQUIVALENCES CAN BE WRITTEN:
F
The minimum gasket load is
M
F
min H G (M , F ) = H G 0 − 2 −
2πa
πa
M
F
bolts
max HG(M,F)
min H G(M,F)
The maximum gasket load is
M
F
max H G (M , F ) = H G 0 + 2 −
2πa
πa
G=2a
No pressure, external bending moment
and external tensile force
IS EQUIVALENT WITH
Feq
F
The equivalence is considered in terms
of “tensile force that gives the reaction on
gasket equal to the minimum gasket load given
by the external loads”:
bolts
HG (Feq )
min H G ( M , F ) = H G ( Feq ) or
HG (Feq )
G=2a
No pressure, equivalent tensile force
IS EQUIVALENT WITH
Feq
F 
 M
H G0 −  2 +
 = H G0 −
2πa 
2πa
 πa
That means:
Feq
M
F
= 2+
, i.e.
2 πa πa
2 πa
2M
4M
Feq = F +
=F+
a
G
Remark: This equivalence is conservative made
and is really true just for a point, not for the entire gasket
circumference.
Feq pressure
F
HG (peq )
The equivalence is considered in terms of “pressure that
gives the same effect as Feq ”, i.e.
bolts
HG (peq )
p eq πa 2 = Feq
or p eq =
G=2a
Equivalent pressure, no external loads
-3-
1 
2M 
F
2M
F+
= 2 + 3
2 
a  πa
πa 
πa
=
G
a=
2
4 F 16M
+
πG 2 πG 3
Understanding the Kellogg Equivalent Pressure Method for piping flanges
GASKET LOAD LIMITS
Fpressure
For the tensile loaded part of the flange, the minimum
gasket load is:
Fpressure + Feq pressure
πa 2 (p + peq )
H G (peq ,p) = H G 0 −
= HG0 −
2π a
2π a
Feq pressure
F
bolts
HG (peq ,p)
HG (peq ,p)
This load must be limited to the value corresponding to
the rating pressure.
G=2a
Internal pressure, equivalent pressure
versus the limit given by the rating pressure
FRATING pressure
F
bolts
H G ( p RATING ) = H G0 −
HG (pRATING )
πa 2 p RATING
2π a
HG (pRATING )
G=2a
The gasket tightness condition means to accept for the gasket load only values that are greater than the
value corresponding to the rating pressure.
The condition H G ( p eq , p) ≥ H G ( p RATING ) means p + p eq ≤ p RATING .
For the compressed loaded part of the flange, the maximum gasket load is:
M
F
max H G ( M , F ) = H G0 + 2 −
, where F > 0 is a tensile force, F < 0 is a compressive force.
2 πa
πa
1
M
F 
This load gives a compressive stress for the gasket:  H G0 + 2 −
 that may be limited to a
b
2 πa 
πa
maximum stress on the gasket. This condition is not covered by limiting p + p eq ≤ p RATING .
-4-