Pressure Equipment Hydrotesting Reasons Pressure Equipment Hydrotesting Reasons.pdf Stress Analysis-Design of Shell and Flat Heads in Pressure Equipment Synopsis of Shell/Flat Heads Stresses & Calculations

1. Synopsis of Shell & Circular Flat Heads equations/calculations in
pressure equipment codes for internal pressure.
Typical pressure equipment mainly comprises cylindrical, ellipsoidal, spherical and flat components. For thin
walled (radius to thickness ratio > 10) component like shells, spherical and ellipsoidal heads the required
thickness for internal pressure is based on a very basic equation or formula where stresses are calculated in
circumferential or tangential and longitudinal or meridional direction.
As an example of thin walled shell under internal pressure the most common equation (Barlow’s formula)
in simplified forms assuming uniform stresses across the wall thickness are:
(Here, S= induced stress or allowable stress, P= pressure, t = thickness, D=diameter, R= Radius, E=joint
efficiency)
Circumferential or hoop Stress =Sh= PD/(2t ) ----(A-1) or PR/t ---- (A-2)
Longitudinal Stress = SL= PD/(4t) ----- (A-3) or PR/(2t )------- (A-4)
Lets compare above to ASME SEC VIII DIV 1 equations:
Circumferential Stresses as per UG-27- (1): t= PR/(SE-0.6P) ----- (B-1)
Longitudinal Stresses as per UG-27 – (2): t= PR/(2SE+0.4P)----(B-2)
Considering the ASME equations are in form of allowable stress to find thickness, we can change equations A-
2 and 4 in the same form as below:
t= PR/ Sh -------- (C-1)
t= PR/ (2SL)-------- (C-2)
Considering for thin wall shells (using E=1), where typically R/t >10, we will have S>>P (i.e.
induced/allowable stress will be significantly greater than internal pressure). Therefore we can assume that
in the denominator the term involving P can be ignored or almost equal to zero in comparison to “S” term so
we have.
t= PR/(S*1-0.6P) = PR/S -----(D-1)
t=PR/(2S*1+0.4P) = PR/(2S) ---- (D-2)
Comparing equation C-1 to D-1 and C-2 to D-2, it is very obvious that the code equations are based on the
basic Barlow’s equations of stresses in shell. However the code equations are adjusted by some factors to
some how capture the uneven stress distribution in a thin walled shell. For a thin walled shell (typically R/t
>10), the difference between the basic Barlow’s equation and the actual code equation may be no greater than
5%.
Using a similar comparison as done above the equations of other codes (i.e. ASME I, ASME B31.1/31.3 etc.) can
be compared to the basic Barlow’s equations and it can easily be seen that the equation are almost equivalent
with some adjustments to code equations and that the difference in results are quite small (<5%) for thin
walled shells.
In all above code equations of the value “S” is allowable stress and considering the stresses in the shell are
induced to internal pressure, which is a primary type of loading therefore the value of allowable stress in the
case will be the be based on the one used for primary membrane stresses, which is the value of “S” listed in
ASME SEC II D (for ASME codes) for the material with no increase by a factor such as 1.2, 1.5 or 3.

2. Now consider a case of a flat heads subjected to pressure loading. Flat heads primarily comes in two
configurations
a) Welded to shells
b) Bolted or clamped to flanges
a) Flat heads welded to shell: There are primarily two types of stresses in the flat heads welded to shell
when subjected to pressure.
I. Primary Bending stresses in the flat head
II. Discontinuity stresses at the flat head to shell junction (secondary stresses)
In most cases for flat heads in codes the only stresses calculated and considered are primary bending
stresses for thickness calculations, while discontinuity stresses (secondary stresses) which are self
limiting are not calculated and some how the code material ductility considerations/requirements, the
factor of design margin on primary membrane/bending stresses, limitation on attachments /specific
attachment details only, compensates for not calculating secondary stresses. For a fatigue analysis
these discontinuity stresses or better in terms of strains may need to be calculated.
As an example here we are considering ASME SEC VIII DIV 1 as a reference code to discuss the case of
flat head. Before going into the code equations, a generalized equation for bending moments can be
derived for flat heads.
Considering simple cases of beams i.e. simply supported at ends (free to rotate) or fixed at ends (no
rotation). For a simply supported beam of length “L” with uniform load distribution “w” per unit
length, the maximum bending moment is in the center and is:
Mbsmax = wL2/8 ---- (E-1)
In case of fixed ends the maximum bending moment is at the ends and is:
Mbfmax = wL2/12 ---- (E-2)
Applying the above analogy to a flat plate, we can consider the flat plate as a beam, which is supported
at its outer periphery, and depending on how much the flat plate is able to rotate at its edges will
determine the maximum bending moment per unit length. If the flat plate is firmly clamped at edges
which means no rotation than it is considered fixed and if lets say welded to a very thin shell which
provided no rotation support/resistance than the flat plate is simply supported at edges where
bending moment and so stresses in the flat head will be more than the one which is welded to thick
walled shell and relatively fixed.

3. To prove the above point lets consider the equations for flat heads using a mathematical/analytical
approach (refer to Roark’s formulas, Table 11.2, case10a & b or refer to Brownell & Young Chapter 6).
For flat head with simply support at edges the maximum moment per unit length is at the center and
is:
Mpsmax= pd2 (3+v)/64 (where p = pressure, d= diameter, v= poison ratio)
Using v=0.3 for typical steels in process equipment above equation becomes:
Mpsmax = 3.3 pd2 /64------- (F-1)
Now for flat head with fixed support at the edges the maximum moment per unit length is at the edges
and is:
Mpfmax= pd2/32 ---------- (F-2)
Now comparing the relative difference of flat heads to that of beams by dividing E-1 by E-2 and F-1 by
F-2 we get
E-1/E-2 = Mbsmax/Mbfmax = 12/8 = 1.5 (simply supported beam have 50% more moment than fixed
case)
F-1/F-2 = Mpsmax/Mpfmax = 3.3 x 32 /64 = 1.65 (simply supported plate have 65% more moment than
fixed case)
The results of beam analogy are giving comparable to plates. Looking at the above results it is clear
that flat plates with minimum rotational restraint at the edges will have maximum bending moment
per unit length. The actual restraint of the plate edge conditions will lie somewhere between these two
extremes of free and fixed. In reality it will never be fully free and nor fully fixed.
Now for welded plate to shells one may consider it a fully fixed type of support however this will not
be conservative as depending on the thickness of the shell element and the type of welded attachment
there may be some rotation at plate to shell junction which will result in relatively higher bending
moment per unit length in the flat plate. Therefore it may be more practical to somehow
adjust/increase the bending moment for fixed value by some factor. Similarly is the case in code
calculations where some constants are used for different type of welded attachment to the shell.
Lets derive the thickness equation for the flat heads and compare it to the ASME SEC VIII DIV 1 code
equations. Before deriving thickness equations it is important to note that pressure equipment codes
have different allowable stresses for material for different type of stresses or loadings. Typically only
one base allowable stress value is listed which is for the primary membrane allowable stress value and
the value for other types of stresses (primary bending, local primary and secondary stresses) is based
on a factor applied to the base allowable value. In case of ASME SEC VIII DIV 1 the base allowable for
primary membrane stress is S (i.e. typically min 2/3 Yield point or tensile strength/3.5), and for
primary bending stress it is 1.5 S (i.e. can reach yield point) and for secondary stresses it is 3S (i.e. can
reach twice yield point).
In case of flat head subjected to internal pressure, the stresses generated in the plate are primary
bending stresses and therefore the allowable will be 1.5S. The reason for using 1.5S is that for plate
failure in bending, the entire cross section to be at a yield stress and this will not happened until the
load is increased above the yield moment of the plate multiplied by the a factor known as shape factor
and the shape factor for a simple rectangular cross section in bending is 1.5.

4. For a plate section thickness “t” where bending moment per unit length is “M”, and the moment of
inertia per unit length as 1/12 (1) t3 :
Bending Stress Sb = M (t/2)/(1/12 t3) = 6M/t2 ------ (G-1)
Taking bending stress as allowable as per ASME SEC VIII DIV 1, which is 1.5S and solving for thickness
we get.
t = (6M/1.5S)1/2 ------- (G-2)
Now considering moment per unit length “M” for simple supported ends and fixed ends case using
equation F-1 & F-2, into equation G-2 as:
t = (6 x Pd2 /(32 x 1.5 S))1/2 = d( (6/(32 x 1.5) x P /S)1/2 ---------- (G-3) (fixed)
t = (6 x 3.3 xPd2 /(64x 1.5 S))1/2 = d( (6 x 3.3)/(64 x 1.5) x P /S)1/2 ---------- (G-4) (simple supported)
The above equation can now be compared to ASME SEC VIII DIV 1 Section UG-34 (c)(2) (1), which is:
t= d (CP/(SE))1/2 ------------ (H-1)
For E=1 seamless head we have:
t= d (CP/S)1/2 ------------ (H-2)
Comparing equation H-2 to G-3/G-4, both are identical except the constant “C”. The constant “C” as per
equation G-3 and G-4 is:
6/(32x 1.5) = 0.125 --- -----------(for fixed support at edges)
(6 x 3.3)/(64 x 1.5) = 0.206 ---- (for simple support at edges)
Now comparing the range we got from 0.125 to 0.206 for constant “C” for welded configuration, when
compared with the constant “C” as per ASME SEC VIII DIV 1, the results is in agreement.

5. In the above we can see constant C is mostly in range of 0.1 to 0.2 except where in some special cases it
is mentioned as 0.33m/0.3, which can be attributed to the fact that for these special geometries the
code may be putting a more stringent limit on stress to possibly reduce discontinuity stresses at plate
to shell junction which may be done by putting a limit on primary bending stresses as less than the
typical value of 1.5S.
It is now also obvious that for most welded flat plates the ASME SEC VIII DIV 1 code is allowing the
plate to reach a stress value of 1.5S, which can be the yield point of material. Looking at the equation
H-1, it may give one an impression that the code is using just a value of S instead of 1.5S, however it is
to be noted that this is not case and the factor 1.5 is already incorporated into the constant “C”. Refer
to the following interpretation.
b) Bolted or clamped to flanges
Apart from welded configuration there two other types of flat head configurations that is typically
used in pressure equipment:
I. Clamped between two surfaces (or flanges)
II. Bolted to flange
I. Clamped between two surfaces:
The flat head in this configuration is ideally considered fixed at the edges. It is also to be noted
that unless the case of welded configuration, in this case the joint is susceptible to leakage if
yielding occurs in the flat plate causing the sealing surfaces to deform. Therefore it is more
appropriate to limit the primary bending stress in the flat plate to less than yield point i.e. S
(2/3 yield point) instead of 1.5S (i.e. yield point) . Using allowable stress of S instead of 1.5S we
can re-arrange equation G-3 above as:
t = (6 x Pd2 /(32 x 1.5 S))1/2 = d( (6/(32 x 1.5) x P /S)1/2 Here “1.5” factor is removed.
t= d( (3/16 x P /S)1/2 -----------(J-1) where simplifying 6/32 = 3/16
Now comparing the above equation J-1 to ASME B31.3 – 304.5.3 used for calculating spectacle
blinds thickness, which is given as:
t= d( (3/16 x P /(SE))1/2

6. Using E=1 (seamless case) t= d ((3/16 x P /S) 1/2 ------------(K-1)
Equation J-1 and K-1 is identical which affirms that the plate bending stresses are kept below
yield point for a leak tight joint.
II. Bolted to Flange:
In this case the flat head is not only subjected to internal pressure but also the operating
bolting loads. The thickness required in this case will obviously be greater than without the
bolting case. To solve for this case we need to find the bending moment per unit length in the
flat plate due to pressure and due to operating bolting loads. It is to be noted that bolted joint
are susceptible to leakage if yielding occurs in the main components and therefore it is more
appropriate in this case to limit the primary bending stress in the flat plate to basic allowable
stress S (i.e. 2/3 of yield) instead of 1.5S (i.e. yield point).
Now to find the bending moment per unit in the flat plate, consider two cases of a beam. One is
analogous to the operating bolting loads where “W” is acting at distance of “c” from the gasket
location while in the other case distributed load is on the beam which is analogous to the
internal pressure acting within the gasket diameter “L”.
Above Analogous to Bolting (Maximum Moment = Wc between length “L”)
Above Analogous to Internal pressure (Maximum moment = wL2/8 at the center)
The total bending moment along the beam simultaneously subjected to above loads can be
easily calculated by superimposing the individual bending moments and the maximum
bending moment will be at the center, which is Wc+ wL2/8.
This above case is equally applicable to the bolted circular flat plates as well. The two
components of the bending moments are:
1) Due to internal pressure on the plate and the plate is simply supported at the gasket
location at diameter “d” since it can rotate at this support point. From equation F-1 we
have
MP = 3.3 pd2 /64
2) Due to operating bolt load which in total is “W” and using the moment arm of “hG” (gasket
to bolt diameter circle) the bending moment is “WhG”, and converting the moment as

7. moment per unit length at point of gasket along the circumference and is equal to Mbolt
=WhG/ (πd).
Using the beam analogy the maximum moment per unit length in the flat plate will be equal to:
Mmax = MP+ Mbolt = 3.3 pd2 /64 + WhG/ (π d)
Using equation G-1, having allowable stress of S instead of 1.5S and solving for “t” we get.
t = (6Mmax/S)1/2
Putting Mmax in above equation we get
t= (6 x 3.3 pd2 /(64 S) + 6 WhG/ (π d S))1/2
t= (0.309 pd2 /S + 1.9 WhG/ (d S))1/2
t= d(0.309 p/S + 1.9 WhG/ (d3 S))1/2 ---------------- (L-1)
The equation in the final form can be compared to ASME SEC VIII DIV 1 – UG-34 (c) (2) (2) which is:
t= d(C p/(SE) + 1.9 WhG/ (d3 SE))1/2
Using E =1 (Seamless Plate) we have
t= d(C p/S + 1.9 WhG/ (d3 S))1/2 --------------------- (M-1)
Comparing equation M-1 to L-1, both are almost identical where the value of “C” as per equation L-1 is
0.309, which is the same as per ASME SEC VIII DIV 1 –Fig UG-34.
There are also interpretations in ASME SEC VIII DIV 1 regarding the use of S instead of 1.5S for
bolted flat heads and the reason being to avoid yielding of plate causing leakage of the bolted joint.
Refer to the following interpretation VIII-1-01-78

8. In summary for a given pressure and dimension, in flat heads the bending stresses depending on the end
support i.e. whether it is bolted/clamped or welded and these primary bending stresses are limited to “1.5S”
for welded configurations and “S” for bolted/camped configuration. In most case for welded configurations
the discontinuity stresses which are secondary stresses at the plate to shell junction, are not calculated.