--------------------------------------- ------------------------------------------------------
  ![](media/image1.png){width="0.875in"   **Fermilab**
  height="0.8194444444444444in"}          

  --------------------------------------- ------------------------------------------------------

**Particle Physics Division**

**Mechanical Department Engineering Note**

# Number: MD-ENG- 088 Date: Sep. xx, 2005

Project Internal Reference:

Project: **ILC**

Title: Vacuum Vessel Engineering Note for 3.9 GHz, 3^rd^ Harmonic

Author(s): Edward Chi

Reviewer(s):

Key Words:

Crymodule, Vacuum, Pressure, FEA, Vessel Shell, Stiffening Rib,

Reinforcement, Flange, Weldment, Up & Downstream,

Abstract Summary:

The vacuum vessel of 3.9 GHz, 3^rd^ harmonic has 42.87" O. D, \~ 82.64"

Length with multiple openings for different connections and

instrumentation. This engineering note is presented extensive

discussion, FEA analysis and calculations for the vessel and nozzle

shells, stiffening rings, multiple openings and reinforcements,
different

flanges, welds and others per Fermilab, ASME and other applicable

codes.

Applicable Codes:

ES&H Manual Chapter 5033, Fermilab.

"Boiler & Pressure Vessel Code" ASME VIII, Div.1

2004 Edition

"Allowable Stress Design", AISC, 9^th^ edition

"Structural Welding Code-Steel", AWS D1.1-90

**EXHIBIT A-1**

**Vacuum Vessel Engineering Note**

**(per Fermilab ES&H Manual Chapter 5033)**

  ------------ -------------- ------- ------------ --------- ---------------
  Prepared by  Edward Chi     Date    08-26-2005   Div/Sec   PPD/MD/ME

                                                             

  Reviewed by                 Date                 Div/Sec   

                                                             

  Div/Sec Head                Date                 Div/Sec   
  ------------ -------------- ------- ------------ --------- ---------------

1\. [Identification and Verification of Compliance]{.underline}

Fill in the Fermilab Engineering Conformance Label information below:

This vessel conforms to Fermilab ES&H Manual Chapter 5033

  -------------------------- --------------------------------------------
  Vessel Title               *3.9 GHz, 3^rd^ Harmonic Vacuum Vessel*

  Vessel Number              

  Vessel Drawing Number      *5520 -- ME - 439317*

  Internal MAWP              *1 psig*

  External MAWP              *14.7 psi*

  Working Temperature Range  *20* ~~^o^~~F *100* ~~^o^~~F

  Design/Manufacturer        *DESY*

  Date of Manufacture        

  Acceptance Date            
  -------------------------- --------------------------------------------

> Director\'s signature (or designee) if vessel is for manned area and
> requires an exception to the provisions of this chapter.

  ----------------------- ----------------------- -----------------------
  **Amendment No.**       **Reviewed by:**        **Date:**

                                                  

                                                  

                                                  
  ----------------------- ----------------------- -----------------------

  ------------------------- ---------------------------------------------
  Laboratory location code  

  Laboratory property       
  number                    

  Purpose of vessel         
  ------------------------- ---------------------------------------------

List all pertinent drawings

  ------------------------------------- ---------------------------------
  Drawing No.                           Location of Original

  *5520 -- ME -- 439317 -1*             

  *5520 -- ME -- 439317 -2*             

  *5520 -- ME - 439262*                 

                                        
  ------------------------------------- ---------------------------------

2\. [Design Verification]{.underline}

Provide design calculations in the Note Appendix.

*See the attached vacuum vessel engineering note #MD-Eng-088*

3\. [System Venting Verification]{.underline}

Can this vessel be pressurized either internally or externally? \[ \]
Yes

\[*X*\] No

####### If "Yes", to what pressure? \_\_\_\_\_\_\_\_\_\_\_\_\_

> List all reliefs and settings. Provide a schematic of the relief
> system components and appropriate calculations or test results to
> prove that the vessel will not be subjected to pressures greater than
> 110% beyond the maximum allowable internal or external pressure.

  -------------------- ------------ ------------- ------------- -------------
  **Manufacturer**     **Relief**   **Pressure    **Flow Rate** **Size**
                                    Setting**                   

                                                                

                                                                

                                                                

                                                                
  -------------------- ------------ ------------- ------------- -------------

4\. [Operating Procedure Section]{.underline}

Is an operating procedure necessary for the safe operation of this
vessel?

> \[ \] Yes \[ *X*\] No *(If \"Yes\", it must be appended)*

Is a testing procedure necessary for the safe acceptance testing
(acceptance

testing) of this vessel?

\[ \] Yes \[ *X*\] No

> *If "Yes", the written procedure must be approved by the division head
> prior to testing and supplied with this Engineering Note.*

5\. [Welding Information]{.underline}

Has the vessel been fabricated in a Fermilab shop? \[ \] Yes \[ *X*\] No

> *If \"Yes,\" append a copy of the welding shop statement of welder
> qualification.*

*The fabrication will be performed by using qualified welder of an*

*outside contractor.*

6\. [Exceptional, Existing, Used and Non-Manned Area
Vessels]{.underline}

Is this vessel or any part thereof in the above categories? \[ \] Yes

\[ *X*\] No

> *If \"Yes\" follow the Engineering Note requirements for documentation
> and append to note.*

**[CONTENTS]{.underline}**

*Section Descriptions Pages*

i\. Exhibit A-1, Fermilab ES&H Manual Chapter 5033 2

1\. Calculation for the Vacuum Vessel Cylindrical Shell Thickness 7

2\. The Permissible Out of Roundness of the Vessel Cylindrical Shell 9

3\. Analysis and Calculations for the Stiffening Rings

3.1 To Define the Downstream End Flange as the Stiffening Ring for the
10

Vessel Shell Under the External Pressure

3.2 To Define the Upstream End Flange as the Stiffening Ring for the 13

Vessel Shell Under the External Pressure

4\. Calculations and Analysis for the Vessel Shell Openings

4.1. Calculations for the Opening of Port Coldmass for the Reinforcement
14

> and Weld Size

4.2. Calculations for the Opening of MC Port (#9) for the Reinforcement
and 19

Weld Size

4.3 Calculations for the Opening of Electronic Port for the 23

> Reinforcement and Weld Size

5\. Calculations and Analysis for the Flanges, Bolts and Welds

5.1 Downstream End Flange, and the Bellow Flange 26

5.2 Upstream End Flange, Slide Flange and the Bellow Flange 36

5.3 Calculations for Flanges #8, #10, #2 and #3 40

6\. Finite Element Analysis (FEA) for the Vessel Weldment, Bottom

Supports and Top Lifting Fixtures 42

7\. Appendix 47

![](media/image2.png){width="5.748611111111111in"
height="4.627777777777778in"}

Figure i.1, [An overall view of the vacuum vessel weldment of 3.9
MHz,]{.underline}

[3^rd^ Harmonic]{.underline}

Reference drawing: ME -- 439317- 1(2)

Overall length = 82.64 in

W~ws~ ≈ 2,264 lbs. the weight of the vessel shell weldment.

Assuming the vessel shell is operating under the room temperature.

Vessel shell material: SA 516-60

Materials for flanges, ports and nozzles: see respective notes and
drawings.

**1. [Calculation for the Vacuum Vessel Cylindrical Shell
Thickness]{.underline}**

*[The Vacuum Vessel Cylindrical Shell:]{.underline}*

Given:

P: 14.7 psi, the external design pressure,

D~o~: 38.00 in, outside diameter of the vessel shell

D~i~: 37.25 in, inside diameter of the vessel shell

t~n~: The nominal vessel shell thickness.

F~a~: 20,000 psi, the max. allowable stress of the shell material under
the operating

temp, -20^o^ F to 100^o^ F, also know as S~v~. for vessel shell material
SA 516-70,

carbon steel, see dwg. #ME-439747 (per Table 1A, part D, section II,
ASME

VIII, 2004 edition)

L~v~: Design length of the vessel between lines of the support.

M.A: Material corrosion and other misc. allowance, \~ 1/16"

Assuming:

L~v~ ≈ 82.64 in (per drawing #ME-439317-01)

Find out the required vessel shell thickness t~n~:

1.  Try t~n~ = 0.31"

L~v~/D~o~ = 82.64" / 42.87" = 1.93

D~o~/t~n~ = 42.87" / 0.31" = 138.29

Per section UG-28, also Fig. G and Fig. CS-2 in Subpart 3 of Section II,
part D,

ASME VIII, Div. 1:

It is found out that:

A = 0.00043

B = 6000

Then:

P~a~ = (4B) / (3D~o~/t~n~)

= (4 x 6000) ÷ (3 x 138.29)

= 57.85 psi

where:

P~a~ : The max. allowable external working pressure, psi

A: Factor determined from Fig. G

B: Factor determined from the applicable material chert in Fig. CS-2.

2.  Try t~r2~ = 0.2

> L~v~/D~o~ = 1.93
>
> D~o~/t~r2~ = 42.87"/0.20" = 214
>
> Find:
>
> A ≈ 0.00023

B = 3300

Then:

P~a2~ = (4B) / (3D~o~/t~n~)

= (4 x 3300) ÷ (3 x 214)

≈ 19.64 psi

3.  Try t~r3~ = 0.175

> L~v~/D~o~ = 1.93
>
> D~o~/t~r3~ = 42.87"/0.175 = 245
>
> Find:
>
> A ≈ 0.00019

B = 2800

Then:

P~a3~ = (4B) / (3D~o~/t~n~)

= (4 x 2800) ÷ (3 x 245)

≈ 15.23 psi

> So, **t~r~** **= 0.175"**
>
> So the vessel shell thickness is required 0.175", using 0.31" as
> vessel shell thickness is above the shell thickness required and it is
> satisfactory.
>
> **2. [The Permissible Out of Roundness of the Vessel Cylindrical
> Shell]{.underline}**

Given:

P: 14.7 psi, the external design pressure,

D~o~: 42.87 in, outside diameter of the vessel shell

D~i~: 42.24 in, inside diameter of the vessel shell

t~n~: 0.31", the nominal vessel shell thickness.

F~a~: 15,000 psi, the max. allowable stress of the shell material at the
operating temp,

also know as S~v~.

Vessel shell material: SA 516-60, carbon steel, see dwg. #ME-439317

L~v~: \~82.64", design length of the vessel between lines of the
support.

The difference between the max. dia and the mini. dia. of the vessel
shell at any cross

section shall not exceed:

1.  x 42.87 in = 0.428 in

> (per section UG-80 (b)(1) of ASME VIII, Div.1)

Per section UG-80(b)(2) and figure UG-80 of ASME VIII, Div.1, it is
found out that

the maximum plus or minus deviation from the true circular form,
measured radially on

the outside or inside of the cylindrical vessel shall not exceed the
maximum permissible

deviation e:

e ≈ 0.65t

= 0.65 x 0.31 in

= 0.202 in

where:

D~o~ / t = 138

L~v~ / D~o~ = 1.93

Per Figure UG-29.2, section UG-29 of ASME VIII, Div.1, it is found out
that:

Arc length ≈ 0.20 D~o~

> = 0.20 x 42.87 in
>
> = 8.574 in
>
> Chord length = 2 x Arc length
>
> = 2 x 8.574 in
>
> = 17.15 in

So, in a chord length of 17.15 in, the maximum plus or minus deviation
from the true

circle form shall not exceed 0.202 in.

**3. [Analysis and Calculation for the Stffening Rings]{.underline}**

1.  **[To Define the Downstream End Flange as the Stiffening Ring for
    the Vessel]{.underline}**

**[Shell Under the External Pressure]{.underline}**

Reference drawings: ME-439317, MD-439240

Givien:

> E = 28 x 10^6^ psi, Modules of elasticity of the flange materials
>
> l~e~ = 1.1 (D~o~t)^0.5^ = 1.1 (42.87" x 0.31")^0.5^
>
> = 4.01", the effective regional length of the vessel shell
>
> F~y~: 25 ksi, min. yield stress of the stiffening material A182-F304L
> under the
>
> operating temperature.

![](media/image3.png){width="5.749305555555556in"
height="4.673611111111111in"}

Figure 3.2, [The cross-section view of the end flange with the vessel
shell, also refer to]{.underline}

[detail L of drawing ME-439317, and drawing MD-439240]{.underline}..

*[To find the required moment of inertia of the combined end flange --
shell cross section about the neutral axis parallel to the axia of the
shell, in^4^, I~s~':]{.underline}*

I~s~' = \[D~o~^2^L~s~(t + A~s~/L~s~)A\] / 10.9 (eq. 3-2)

(per UG-29, ASME VIII, Div.1)

Where:

A~sd~: Cross-sectional area of the downstream end flange stiffening
area, in^2^

≈ (1.57 x 3.74) -- (0.79 X 1.57) in^2^ = 4.6315 in^2^, see figure 3.2
and MD-439240

L~sd~: Distance between the stiffening ring and downstream end flange,
\~82.64 in

( See drawing ME-439317)

B: Factor determined from the applicable chart,

= 0.75 \[PD~o~ ÷ (t + As/Ls)\]

= 0.75 \[14.7 psi x 42.87 in ÷ (0.31 in + 4.6315 in^2^/82.64 in)\]

= 1291

Per figure CS-2 in Subpart 3 of section II, Part D, where B = 1084
falling below the left end of the material/temperature curve, so:

A = 2B/E

= 2 x 1291 / (28 x 10^6^)

= 0.0000922

(per step 5 of UG-29, ASME VIII, Div. 1)

Then the Eq. 3-2 becomes:

I~s~' = {\[42.87^2^ x 82.64 x (0.31 + 4.6315/82.64) x 0.0000922\] /
10.9}in^4^

= ((151,879 x 0.366 x 0.0000922) /10.9) in^4^

= 0.4702 in^4^

*[To find the combined moment of inertia of the cross section area of
the stiffening ring and shell]{.underline}*

It is found that from the figure 3.2:

A~1~ = 2.79 in x 0.31 in = 0.8649 in^2^

A~2~ ≈ (1.57 x 3.74) -- (0.79 X 1.57) in^2^ = 4.6315 in^2^

> y~1~ = 0.155 in
>
> y~2~ = 2.18 in
>
> y = (A~1~y~1~ + A~2~y~2~) ÷ (A~1~ + A~2~)
>
> = 1.8613 in
>
> dy~1~^2^ = (y -- y~1~)^2^ = 2.9115 in^2^
>
> dy~2~^2^ = (y -- y~2~)^2^ = 0.1016 in^2^
>
> I~1~ = 0.0069 in^4^

I~2~ = 6.8444 in^4^

> I = ∑I~i~ + ∑A~i~dy~i~^2^
>
> = (6.8513 + 2.9887) in^4^
>
> = [9.84 in^4^]{.underline} \> I'= 0.4702 in^4^

Since the requirement moment of inertia I' is less than the value of I
provided by the combined moment inertia of the end flange(downstream)
and vessel shell, so the downstream end flange is satisfactory and it
can be treated as a support of the vessel.

2.  **[To Define the Upstream End Flange as the Stiffening Ring for the
    Vessel Shell Under the External Pressure]{.underline}**

Reference drawings: Detail K of drawing ME-439317-2,

MD- 439240

L~su~: Distance between the stiffening ring and upstream end flange, \~
82.64 in = L~sd~

A~su~: ≈ (1.57 x 3.74) -- (0.79 X 1.57) in^2^ = 4.6315 in^2^ = A~sd~

Since using the same flange at the upstream end and the downstream end,
also, L~su~ = L~sd~; A~su~ = A~sd,~ D~ou~ = D~od~, t~u~ = t~d~, from the
calculation steps of section 3.2, we can conclude that the requirement
moment of inertia I~u~' = I~d~', the available combined moment inertia
I~u~ = I~d~, that leads to the conclusion:

I~u~ = I~d~ = [9.84 in^4^]{.underline} \> I~u~' = I~d~'= 0.4702 in^4^

So the upstream end flange is satisfactory and it can be treated as a
support of the vessel.

**4. [Calculations and Analysis for the Vessel Shell
Openings]{.underline}**

1.  **[Calculations for the Opening of Coldmass Port (#3) for the
    Reinforcement and]{.underline}**

> **[Weld Size]{.underline}**

*[The main vessel shell]{.underline}*:

(ref. drawings: ME-439317, ME-4392620, MD-439252)

P: 14.7 psi, external design pressure

D~o~: 42.87 in, outside dia. Of the vessel shell.

D~i~: 42.24, inside dia. of the vessel shell.

t: 0.31 in, the nominal pipe wall thickness.

F~a~: 15,000 psi, the max. allowable stress of the shell material at the
operating temp,

also know as S~v~. at the operating temp., -20^0^ F to 100^0^ F (per
Table 1A, part D,

Section II, ASME Boiler & Pressure Vessel Code 1995 edition).

Vessel shell material: SA 516, grade 60, carbon steel (see ME-439262)

L~v~: Design length of the vessel section between lines of support (see
Page UG-28(b)),

≈ 82.64"

Find out the required vessel shell thickness t~r~:

1\. Try t~n~ = 0.31"

L~v~/D~o~ = 82.64" / 42.87" = 1.93

D~o~/t~n~ = 42.87" / 0.31" = 138.29

Per section UG-28, also Fig. G and Fig. CS-2 in Subpart 3 of Section II,
part D,

ASME VIII, Div. 1:

It is found out that:

A = 0.00043

B = 6000

Then:

P~a~ = (4B) / (3D~o~/t~n~)

= (4 x 6000) ÷ (3 x 138.29)

= 57.85 psi

where:

P~a~ : The max. allowable external working pressure, psi

A: Factor determined from Fig. G

B: Factor determined from the applicable material chert in Fig. CS-2.

2.  Try t~r2~ = 0.2

> L~v~/D~o~ = 1.93
>
> D~o~/t~r2~ = 42.87"/0.20" = 214
>
> Find:
>
> A ≈ 0.00023

B = 3300

Then:

P~a2~ = (4B) / (3D~o~/t~n~)

= (4 x 3300) ÷ (3 x 214)

≈ 19.64 psi

3.  Try t~r3~ = 0.175

> L~v~/D~o~ = 1.93
>
> D~o~/t~r3~ = 42.87"/0.175 = 245
>
> Find:
>
> A ≈ 0.00019

B = 2800

Then:

P~a3~ = (4B) / (3D~o~/t~n~)

= (4 x 2800) ÷ (3 x 245)

≈ 15.23 psi

> So, **t~r~** **= 0.175"**

*[The coldmass port:]{.underline}*

d~o~: 25.197", outside dia. of the port.

D~i~: 15.512", inside dia . of the port.

t~n~: 4.842", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the
operating temp.

Material: SA 182- 304L stainless steel pipe, F~a~ = S~n~ = 16,700 psi at
the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: ME-439317, MD-439251, MD-439252, and also as shown
from

figure 4.1.

![](media/image4.png){width="4.434027777777778in"
height="4.2027777777777775in"}

Figure 4.1, [3D view of the coldmass port #3 and the vessel
shell]{.underline}

Find out the thickness required:

1.  Assuming L~1~ = 5.49 in, t~r1~ = 0.375 in, then:

> L~1~/d~o~ = 5.49/25.197 = 0.2179
>
> d~o~/t~r1~ = 25.197/0.375 = 67
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.0165
>
> B = 13800
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 13800)/(3 x 67) = 275 psi \> P

2.  Try: L~2~ = 5.49 in, t~r2~ = 0.0625 in

> L~2~/d~o~ = 0.2179
>
> d~o~/t~r2~ = 403
>
> get:
>
> A = 0.00083
>
> B = 8250
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8250)/(3 x 403) = 27.30 psi \> P

3.  Try: L~3~ = 6.64 in, t~r3~ = 0.0425 in

> L~3~/d~o~ = 0.2179
>
> d~o~/t~r3~ = 25.197/0.0425 = 593
>
> get:
>
> A = 0.000475
>
> B = 6750
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 6750)/(3 x 593) = 15.18 psi ≈ P
>
> So: **t~rn~** **= t~r3~ = 0.0425 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 and Fig. UW-16.1(p) of ASME VIII, Div.1.

t~o~: the throat dimension of the outer attachment weld, ≥ 0.5 t~min~.

t~i:~ the throat dimension of the inside of the shell cutout, ≥
0.7t~min~

t~o~ = 0.5 t~min~ = 0.5 x 0.31" = 0.155", **so the fillet leg size is
0.22"**

t~i~ = 0.7 t~min~ = 0.7 x 0.31" =0.217", **so the fillet leg size is
0.31"**

(See details J & K of dwg. ME-439317-1 as the references)

So the sizes of the weld are satisfactory.

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~)

= 0.5 x (16.299" x 0.175" x 1.0 +2 x 4.842" x 0.25" x 1.0 x 0)

= **1.4262 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 16.299 in (per drawing ME-439262)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ =1.0 for nozzle wall abutting the vessel wall as shown in figure
UG-40(j),

(k), (n), (o) and (p).

f~r2~ = S~n~ / S~v~ = 16.7 / 15 = 1.113

*[Area of reinforcement available A~a~]{.underline}:*

A~1~: Larger of the following calculated value: (area available in
shell):

= d(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 16.299" x (1 x 0.31" -- 1.0 x 0.175") -- 0

= 2.2004 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.31" + 4.842") x (1.0 x 0.31" -- 1.0 x 0.175") -- 0

= 1.391 in^2^

So pick **A~1~ = 2.2004 in^2^**

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (4.842 in -- 0.0425 in) x 0.31 in x 1.0

= 7.4392 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (4.842 in -- 0.0425 in) x 1.113 x 4.842 in

= 129.33 in^2^

So pick **A~2~ = 7.4392 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.22^2^ in^2^ x 1.0

=**0.0484 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.31^2^ in^2^ x 1.0

= **0.0961 in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (2.2004 + 7.4392 + 0 + 0.0484 + 0.0961) in^2^

= **9.784 in^2^** \> A~r~ = 1.4262 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be also true for coldmass port
opening #2 (see drawing MD-439251), this is because: all basic data such
as: d~o,~ D~i~, L, t~o~, t~i,~ S~n~ of coldmass #2 are the same as
coldmass #3. So, per section UG-37 of ASME VIII, Div.1, the #3 opening
is adequately reinforced under current design.

1.  **[Calculations for the Opening of MC (main coupler) Port (#9) for
    the]{.underline}**

> **[Reinforcement and Weld Size]{.underline}**

[*The main vessel shel*l]{.underline}:

From section 4.1, it is found that **t~r~** **= 0.175"**

[*The MC port*:]{.underline}

d~o~: 11.339", outside dia. of the port.

D~i~: 10.945", inside dia . of the port.

t~n~: 0.197", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the
operating temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: ME-439317-2, and also as shown from figure 4.2.

![](media/image5.png){width="2.8833333333333333in"
height="2.8118055555555554in"}![](media/image6.png){width="2.9916666666666667in"
height="2.783333333333333in"}

Figure 4.2, [The partial 2D and 3D views of the MC port from
ME-439317]{.underline}

Find out the wall thickness required:

1.  Assuming L~1~ = 8.57 in, t~r1~ = 0.125 in, then:

> L~1~/d~o~ = 8.57/11.339 =0.7558
>
> d~o~/t~r1~ = 11.339/0.125 = 91
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.00225
>
> B = 10500
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 10500)/(3 x 91) = 153 psi \> P

2.  Try: L~2~ = 8.54 in, t~r2~ = 0.0625 in

> L~2~/d~o~ = 0.7558
>
> d~o~/t~r2~ = 181
>
> get:
>
> A = 0.00075
>
> B = 8050
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8050)/(3 x 181) = 59.3 psi \> P

Try: L~3~ = 8.54 in, t~r3~ = 0.033 in

> L~2~/d~o~ = 0.7558
>
> d~o~/t~r3~ = 344
>
> get:
>
> A = 0.000280
>
> B = 3950
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 3950)/(3 x 344) = 15.31 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.033 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 and Fig. UW-16.1© of ASME VIII, Div.1.

t~min~: 0.197"

t~c:~ ≥ 0.7t~min~ = 0.138", **so the fillet leg size is 0.195"**

Full penetration weld (bevel groove) plus the fillet weld t~c.~

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~)

= 0.5 x (11.339" x 0.175" x 1.0 +2 x 0.197" x 0.175" x 1.0 x 0.053)

= **0.994 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 11.339 in (per drawing ME-439317)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 15 =0.947

*[Area of reinforcement available A~a~]{.underline}*:

A~1~: Larger of the following calculated value: (area available in
shell):

= d(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 11.339" x (1 x 0.31" -- 1.0 x 0.175") -- 2 x 0.197 (1.0 x 0.31 -- 1.0
x 0.175) x

0.053

= (1.5308 -- 0.0028) in^2^

= 1.528 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.31" + 0.197") x (1.0 x 0.31" -- 1.0 x 0.175") -- 2 x 0.197 (1.0 x
0.31 -- 1.0 x

0.175) x 0.053

= (0.1825 -- 0.0028) in^2^

= 0.1341 in^2^

So pick **A~1~ = 1.528 in^2^**

A~2~: Smaller of the following:

= 5 (t~n~-t~rn~)tf~r2~

= 5 x (0.197 in -- 0.033 in) x 0.31 in x 0.947

= 0.2407 in^2^

or = 5 (t~n~-t~rn~) f~r2~t~n~

= 5 x (0.197 in -- 0.033 in) x 0.947 x 0.197 in

= 0.153 in^2^

So pick **A~2~ = 0.153 in^2^**

A~3~: Area available in inward nozzle, use the smallest value:

Since there is no inward nozzle

So pick **A~3~ = 0.0 in^2^**

A~41~: Outward nozzle weld

= leg^2^ f~r2~

= 0.195^2^ in^2^ x 0.947

=**0.036 in^2^**

A~43~: Inward nozzle weld

= leg^2^ f~r2~

= 0.0^2^ in^2^ x 1.0

= **0.0 in^2^**

Then:

A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

= (1.528 + 0.153 + 0 + 0.036 + 0.0) in^2^

= **1.717 in^2^** \> A~r~ = 0.994 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be true for all 2 openings (#9
openings, see drawings ME-439317 & ME-439262)

> **4.3 [Calculations for the Opening of Electronic Port (#11
> opening)]{.underline}**
>
> **[for the Reinforcement and Weld Size]{.underline}**

*[The main shell (Tube #9)]{.underline}*

From section 4.2, it is found that: **t~r~ = 0.033 in**

*[The instrumentation port:]{.underline}*

d~o~: 3.50", outside dia. of the port.

D~i~: 3.334", inside dia . of the port.

t~n~: 0.083", wall thickness of the port.

t~rn~: required thickness of the port.

L: length of projection defining the thickness portion of integral
reinforcement of the

port beyond the outside surface of the vessel wall.

E: 28 x 10^6^ psi, modulus of elasticity of the material at the opening
temp.

Material: SA 249-TP304L stainless steel pipe, F~a~ = S~n~ = 14,200 psi
at the operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

Reference drawings: section D-D, ME-439317, and also as shown from
figure 4.2.

Find out the wall thickness requires:

> Assuming L~1~ = 2.0075 in, t~r1~ = 0.083 in, then:

where: L~1~ = 7.48 -- (11.339/2 -- 0.197) = 2.0075"

> L~1~/d~o~ = 2.0075/3.50 =0.5736
>
> d~o~/t~r1~ = 3.50/0.083 = 42
>
> From Fig. G and Fig. HA-3 of subpart 3 of Sect. II, part D:
>
> A = 0.009
>
> B = 13200
>
> P~a~ = (4B)/(3(d~o~/t~r1~)) = (4 x 13200)/(3 x 42) = 419 psi \> P

Try: L~2~ = 2.0075 in, t~r2~ = 0.015625 in

> L~2~/d~o~ = 0.5736
>
> d~o~/t~r2~ = 224
>
> get:
>
> A = 0.0007
>
> B = 8000
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 8000)/(3 x 224) = 47.62 psi \> P
>
> Try: L~3~ = 2.0075 in, t~r3~ = 0.0094 in
>
> L~3~/d~o~ = 0.5732
>
> d~o~/t~r3~ = 372
>
> get:
>
> A = 0.0003
>
> B = 4250
>
> P~a~ = (4B)/(3(d~o~/t~r2~)) = (4 x 4250)/(3 x 372) = 15.23 psi ≈ P

So: **t~rn~** **= t~r3~ = 0.0094 in**

*[Size of the weld required]{.underline}*:

Per UG-37, UW-16 and Fig. UW-16.1(c) of ASME VIII, Div.1.

t~min~: 0.083"

t~c:~ ≥ 0.7t~min~ = 0.056", **so the fillet leg size is 0.08"**

Full penetration weld (bevel groove) plus the fillet weld t~c.~

*[Area of the reinforcement required A~r~]{.underline}*:

Per section UG-37(d)(1):

A~r~ = 0.5(dt~r~F + 2t~n~t~r~F(1-f~r1~)

= 0.5 x (3.5" x 0.033" x 1.0 + 2 x 0.083" x 0.033" x 1.0 x 0)

= **0.0578 in^2^**

Where:

d: Finished diameter of circle opening, see Fig. UG-37.1 of UG-37.

D = 11.339 in (per drawing ME-439262)

F: Correction factor

f~r~ : Strength reduction factor, not greater than 1.0

f~r1~ = S~n~ / S~v~ for nozzle wall inserted through the vessel wall

f~r2~ = S~n~ / S~v~ = 14.2 / 14.2 = 1.0

*[Area of reinforcement available A~a~]{.underline}*:

A~1~: Larger of the following calculated value: (area available in
shell):

= d(E~1~t -- Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 3.50" x (1 x 0.197" -- 1.0 x 0.033") -- 2 x 0.083 (1.0 x 0.197 -- 1.0
x 0.033) x 0

= 0.574 in^2^

or = 2(t + t~n~)(E~1~t --Ft~r~)-2t~n~(E~1~t -- Ft~r~) x (1-f~r1~)

= 2 (0.197" + 0.083") x (1.0 x 0.197" -- 1.0 x 0.033") -- 2 x 0.197 (1.0
x 0.375 -- 1.0

x 0.25) x 0.0

= (2 x 0.28 x 0.164) in^2^

= 0.0918 in^2^

So pick **A~1~ = 0.574 in^2^**

Since A~1~ = 0.574 in^2^

Also: A~a~ = A~1~ + A~2~ + A~3~ + A~41~ + A~43~

That leads to:

A~a~ \> A~r~ = 0.0578 in^2^

# **Per Fig. UG-37.1, section UG-37, the opening is adequately reinforced under the current designated boundary conditions.**

The analysis and conclusion should be true for all 2 electronic ports
(#11 openings, see section D-D, drawing ME-439317 and Figure 4.2 of Page
19).

**5. [Analysis and Calculations for the Flanges,Their Bolts and
Welds]{.underline}**

**5.1 [Downstream End Flange and the Bellow Flange]{.underline}**

References: ME-439317-2, MD-439240

Endcap (module 2-3), #0988348/0, DESY -- MKS1

and Figure 5.1

![](media/image7.png){width="5.502777777777778in"
height="4.988888888888889in"}

Figure 5.1, [Downstream end flanges with the below flange]{.underline}

Figure 5.1 is the assembly configuration of the downstream flange
connecting with the below flange per DESY's layout "Endcap (module 2-3)"

It was found that:

A = 50.35" (dia.)

B~1~ = 47.34" (dia.)

B~2~ = 42.24" (dia.)

C = 49.21" (dia.)

G = 47.889" (dia.)

T~1~ = 1.575"

T~2~ = 1.575"

P = 14.7 psi

n = 24

E~1~ = E~2~ = 2.8 x 10^7^ psi

Material: SA 182-F304L stainless steel, F~a~ = S~f~= 16,700 psi at the
operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

assuming bolt spec: ½ - 13 UNC

A~b~ = 24 x 0.1257 in^2^ = 3.0168 in^2^

The cross-sectional area of the bolts using the root dia. (0.4001") of
the

thread in sq. in.

Per section Y-5.1, Y-5.2, figure Y-5.1.2 of Appendix Y,

Section 2-4, figure 2-4 of Appendix 2 of ASME VIII:

B~2~ \> ½ C,

So it is defined as: Class 2, Category 3 flange

*[Flange Analysis]{.underline}*:

The moment due to the pressure M'~p~:

M'~p~ = H~D~'h~D~' + H~T~'h~T~' (per Y-6.2(3), Appendix Y, ASME VIII)

H~D~'= 0.785 B~2~^2^ P

= 0.785 x 42.24^2^ in^2^ x 14.7 psi

= 20,589 lbs.

H~T~'= 0.785P (B~1~^2^ --B~2~^2^)

= 0.785 x 14.7 psi (47.34^2^ -- 42.24^2^) in^2^

= 5,272 lbs.

h~D~' = (B~1~ -- B~2~) ÷ 2

= (47.34 -- 42.24) in ÷ 2

= 2.55 in

h~T~' = (B~1~ -- B~2~) ÷ 4

= (47.34 -- 42.24) in ÷ 4

= 1.275 in

So: M'~p~ = (20,589 x 2.55) lbs-in + (5,272 x 1.275) lbs-in

= 59,224 lbs-in

= C~5~

(per section Y-6.2(4)(a), Appendix Y, ASME VIII, Div.1)

Let: C~6~ = 0.829 / log(B~1~/B~2~)

= 0.829 ÷ 0.0495

= 16.747

(per section Y-6.2(4)(b), Appendix Y, ASME VIII, Div.1)

C~1~ = \[1 -- 2.095J~s~ log(A/B~1~)\] ÷ (- C~6~ -- 1.738 J~s~)

(per section Y-6.2(4)(c), Appendix Y, ASME VIII, Div.1)

Where:

J~s~ = (1/B~1~) \[2h~D~/β + h~C~/α\] + π ґ~B~

h~D~ = (C -- B) / 2

= (49.21 -- 47.34) / 2 in

= 0.935 in

h~C~ = (A -- C) / 2

= (50.35 -- 49.21) / 2 in

= 0.57 in

β = ( C + B)/ 2B

= (49.21 + 47.34) / (2 x 47.34)

= 1.02

α = ( A + C)/ 2B

= (50.35 + 49.21) / (2 x 47.34)

= 1.052

ґ~B~ = (1/n)\[4/(1 -- ($\overline{A}\overline{R}$)^2^)^1/2^$$tan^-1^((1+
$\overline{A}\overline{R}$)/(1 - $\overline{A}\overline{R}$))^1/2^ -- π
-2$\overline{A}\overline{R}$\]

where:

$\overline{A}\overline{R}$= nD/ πC

= (24 x 0.551") / π 49.21"

= 0.0855

D: the bolt hole diameter

ґ~B~ = (1/24)\[4/(1 -- (0.0855)^2^)^1/2^$$tan^-1^((1+ 0.0855)/(1 --
0.0855))^1/2^ -- π --

2 x 0.0855\]

= 0.041667 x (4.0147 x 0.8282- π -- 0.171)

= 0.0005159

J~s~ = (1/47.34") \[2 x 0.935/1.02 + 0.57/1.052\] + π x 0.0005159

= 0.0211 x (1.8333 + 0.5418) + 0.0016

= 0.0517

So:

C~1~ = \[1 -- 2.095 x 0.0517 x log(50.35/47.34)\] ÷ ( - 16.747 -- 1.738
x 0.0517)

= ( 1 -- 0.0029 ) ÷ (-16.8369)

= -0.0592

C~2~ = (1.738J~p~M~p~ -- C~5~C~6~) ÷ (- C~6~ -- 1.738 J~s~)

(per section Y-6.2(4)(d), Appendix Y, ASME VIII, Div.1)

where:

J~p~ = (1/B~1~) \[h~D~/β + h~C~/α\] + π ґ~B~

= (1/47.34") \[0.935/1.02 + 0.57/1.052\] + π x 0.0005159

= 0.0211 x (0.9167 + 0.5418) + 0.0016

= 0.0324

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~ (per sectionY-3, Appendix Y, ASME
VIII)

H~D~ = 0.785 B~1~^2^ P

= 0.785 x 47.34^2^ in^2^ x 14.7 psi

= 25,861 lbs.

H = 0.785G^2^P

= 0.785 x 47.889^2^ in^2^ x 14.7 psi

= 26,464 lbs.

H~T~ = H - H~D~

= 26,464 -- 25,861

= 603 lbs.

h~G~ = (C -- G)/2

= (49.21 -- 47.889) in / 2

= 0.6605 in

h~D~ = 0.935 in

h~T~ = (h~D~ + h~G~) / 2

= 0.7978 in

H~G~ = W -- H, gasket load

(per section 2-3, appendix 2, ASME VIII)

W = H + H~p,~

flange design load for the operating condition

(per section 2-5(c)(1), appendix 2, ASME VIII)

which leads to:

H~G~ = W -- H

= H + H~p~ -- H

= H~p~

= 2b x 3.14GmP

= 2 x 0.281 in x 3.14 x 47.889 in x 3.5 x 14.7 psi

= 4,348 lbs.

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~

= (25,861 x 0.935 + 603 x 0.7978 + 4,348 x 0.6605) lbs-in

= (24,180 + 481 + 2,872) lbs-in

= 27,533 lbs-in

C~2~ = (1.738J~p~M~p~ -- C~5~C~6~) ÷ (- C~6~ -- 1.738 J~s~)

= (1.738 x 0.0324 x 27,533 -- 16.747 x 59,244) lbs-in ÷ (-16.747 --
1.738 x

0.0517)

= 990,609 lbs-in / (- 16.8369)

= -58,836 lbs-in

For the rigid body rotation of the flange times E\*:

E~1~\* θ~rbi~ = X(C~4~ -- C~2~) ÷ (1.206 log(A/B~1~) --XC~3~ -- (1 --
X)C~1~)

where:

C~3~ = - (0.575 -- 1.206 Js log(A/B~1~)) ÷ (Js + t~1~^3^/F~1~')

= 0

C4 = - (J~p~M~p~) ÷ (Js + t~1~^3^/F~1~')

= 0

$\because$ F~1~'= 0 (per section Y-3(6c), Appendix Y, ASME VIII, Div.1)

X = E~1~\* / (E~1~\* + E~2~\*)

E~1~\* = E~1~ t~1~^3^

E~2~\* = E~2~ t~2~^3^

$\because$ E~1~ = E~2~

$\therefore$ X = E~1~ t~1~^3^ / (E~1~ t~1~^3^ + E~2~ t~2~^3^ )

= t~1~^3^ / (t~1~^3^ + t~2~^3^ )

= 3.907 / (3.907 + 3.907)

= 0.50

E~1~\* θ~rbi~ = 0.50( 0 + 58,836) ÷ \[(1.206 x 0.0268) --(0.50 x 0.0) --
(1 --

0.50) x -0.0592)\]

= 0.50 x (+58,836) lbs-in ÷ (0.0323 -- 0.0 + 0.0296)

= 29,418 lbs-in ÷ 0.0619

= 475,250 lbs-in

E~2~\* θ~rbii~ = - E~1~\* θ~rbi~ (E~2~\* / (E~1~\*)

= - (475,250 x 1.0) lbs-in

= - 475,250 lbs-in

*Total Flange Moment at Diameter B~1~*:

M~s1~ = C~3~ (E~1~\* θ~rbi~) + C~4~

= 0.0

M~s2~ = C~1~ (E~2~\* θ~rbii~) + C~2~

= - 0.0592 x - 475,250 lbs-in - 58,836 lbs-in

= - 30,701 lbs-in

*Unbalanced Flange Moment at Diameter B~1~:*

M~u1~ = 1.206 (E~1~\* θ~rbi~) log(A/B~1~)

= 1.206 x 475,250 lbs-in x 0.0268

= 15,360 lbs-in

M~u2~ = 1.206 (E~2~\* θ~rbii~) log(A/B~1~)

= 1.206 x (- 475,250 lbs-in) x 0.0268

= -15,360 lbs-in

*Balanced Flange Moment at Diameter B~1~:*

M~b1~ = M~s1~ - M~u1~

= 0.0 lbs-in - (15,360 lbs-in)

= -15,360 lbs-in

M~b2~ = M~s2~ - M~u2~

= -30,701 lbs-in + 15,360 lbs-in

= -15,341 lbs-in

*Slope of Flange at Diameter B~1~ times E*:

E~1~ θ~Bi~ = (5.46/π t~1~^3^) (J~s~M~b1~ + J~p~M~p~) + E~1~\* θ~rbi~ /
t~1~^3^

= 0.4448 in^-3^ x (0.0517 x (-- 15,360 lbs-in) + 0.0324 x 27,533 lbs-in)
+

475,250 lbs-in / 3.907 in^3^)

= (44 + 121,641) psi

= 121,685 psi

E~2~ θ~Bii~ = (5.46/π t~2~^3^) (J~s~M~b2~ + J~p~M~p~) + E~2~\* θ~rbii~ /
t~2~^3^

(per section Y-6.2(e), Appendix Y, ASME VIII, Div.1)

= \[0.4448 x (0.0517 x (-15,341) + 0.0324 x 27,533) - 475,250 / 3.907\]
psi

= (44 -- 121,641) psi

= -121,597 psi

*Contact Force between Flange at h~c~:*

H~c~ = (M~p~ + M~b1~) / h~c~

= (27,533 - 15,360) lbs-in / 0.57 in

= 12,173 lbs.

*Bolt Load at Operating Conditions:*

W~m1~ = H + H~G~ + H~C~

= (26,464 + 4,348 + 12,173) lbs

= 42,985 lbs.

*Operating Bolt Stress f~b~*:

f~b~ = W~m1~ /A~b1~

= 42,985 lbs / 3.0168 in^2^ (See page 52, section 5.1 of the note)

= [14,249 psi]{.underline}

*Design Pre-stress in Bolts:*

S~i~ = f~b~ -- 1.159 h~c~^2^ (M~p~ + M~b1~) / (2(1 -- X)a t~1~^3^ l
r~e1~B~1~)

where:

r~e1:~ ≈ 1.0, elasticity factor, the ratio between E~flange~ and E~bolt~

a = (A + C) / 2B~1~

= (50.35 + 49.21) in / 2 x 47.34 in

= 1.052 (shape factor)

l: = t~1~ + t~2~ + 0.5d~b~ (calculated strain length of the bolt)

= (1.575 + 1.575 + 0.25) in

= 3.40 in

S~i~ = 14,249 psi-- 1.159 x 0.57^2^ in^2^ x (27,533 -- 15,360)lbs-in /
(2(1 -- 0.50) x

1.052 x 1.575^3^ in^3^ x 3.40 in x 1.0 x 47.34 in)

= 14,249 psi -- 4,584 lbs-in^3^ / 662 in^5^

= 14,242 psi

*Redial Stress in Flange I @ Bolt Circle*:

S~R1~ = 6(M~p~ + M~S1~) / t~1~^2^(πC --nD)

= 6 x ( 27,533 -0) lbs-in / 1.575^2^ in^2^ x (π 49.21 in -- 24 x 0.551
in)

= 165,198 lbs-in / 350.7 in^3^

= 471 psi \< S~n~ = 16,700 psi

*Tangential Stress in Flange I @ Inside Diameter*:

S~t1~ = t~1~ E~1~ θ~Bi~ / B~1~ + (2Ft~1~ Z/(ho + Ft~1~)
-1.8)(M~s1~/(πB~1~t~1~^2^)

= 1.575 in x 121,685 psi / 47.34 in

= 4,048 psi \< S~n~ = 16,700 psi

*Tangential Flange Stress Adjacent to Central Nozzle*:

S~t2n~ = Y(M'~p~ + M~S2~) / t~2~^2^ B~2~

where:

Y = (1/(K-1))\[0.66845 + 5.71690 K^2^log~10~K/(K^2^ -1)\]

(per figure 2-7.1, Appendix 2, ASME VIII, Div. 1)

K = A/B

= 50.35 / 42.24

= 1.192

Y = (1/(1.192-1))\[0.66845 + 5.71690 x 1.4209 x log~10~K/(K^2^ -1)\]

= 5.2083 x (0.6685 + 5.7169 x 0.2575)

= 11.1489

$\therefore$ S~t2n~ = 11.1489 x (59,224 -- 30,701) lbs-in / (1.575^2^ x
42.24) in^3^

= 3,035 psi \< S~n~ = 16,700 psi

*Longitudinal Hub Stress in Flange 1:*

S~h1~ = 0

(per equation (34c), section Y-6.3, appendix Y, ASME VIII, Div.1)

*Radial Stress in Flange 2 @ bolt circle*:

S~r2b~ = 0

*Radial Stress in Flange 2 @ Diameter B~1~*:

S~r2B1~ = 0

*Tangential Stress at Center of Flange 2*:

S~t2B2~ = (0.3094 PB~1~^2^ / t~2~^2^) -- (6 M~S2~ / πB~1~ t~2~^2^)

= (0.3094 x 14.7 psi x 47.34^2^ in^2^ / 1.575^2^ in^2^ )

-   ( 6 x (-30,701 lbs-in) / π 47.34 in x 1.575^2^ in^2^ )

= 4,109 psi + 499 psi

= 4,608 psi \< S~n~ = 16,700 psi

*[Welds Analysis and Calculations:]{.underline}*

See UW-13 (e)(2), Fig. UW-13.2, section 2-3 and Fig. 2-4(4a), Appendix 2
of

ASME VIII, Div.1. for the related codes and the physical representation
of the

nomenclatures.

Per above codes, it is required that the throat of the fillet c =
0.7t~min~. = 0.7 x 0.31"

= 0.22",

So the fillet leg size is \~ 0.31" as shown in figure 5.1.

It is found that:

t~p~= 0.53" \> 0.25"

where is the smallest of t~n,~ t~x~ and 0.25" is 0.25"

c = t~n~ = 0.31" = leg of the fillet

b = 0.16" + 2 x tan 15^o^ x 0.38"

= 0.36"

a = 0.38" + 0.31"- 0.06"

= 0.63"

$\therefore$ a + b = 0.99" ≥ 3t~n~ = 0.93"

The weld size design for the end flange is satisfactory.

*[Also run a weld strength calculation for the noted location as the
2^nd^ analysis]{.underline}*

*[approach]{.underline}*:

Radial pressure load F~rp~:

F~rp~ = P L~sd~

= 14.7 psi x 82.84 in

= 1,215 lbs/in

Shear flow due to radial shear load F~sr~:

F~sr~ = VQ/I

where:

Radial shear load V:

V = 0.01 P L~sd~ D~o~

= 0.01 x 14.7 psi x 82.64 in x 42.87 in

= 521 lbs.

The 1^st^ moment of area Q:

Q = 2.79 in x 0.31 in x (1.8613 -- 0.155) in

= 1.4758 in^3^

I = 9.84 in^4^

F~sr~ = VQ/I

= 521 lbs. x 1.4758 in^3^ / 9.84 in^4^

= 78 lbs/in

The required resultant weld load F~rw~:

F~rw~ = ( 1,215^2^ + 78^2^)^1/2^ lbs/in

= 1,218 lbs/in

The welds for connecting the flange and the vessel shell is single
groove weld inside

and intermittent fillet outside, the lowest allowable weld stress F~aw~:

F~aw~ = 0.60 S~n~ = 0.60 x 16.7 ksi

= 10.02 ksi

The requires calculated minimum weld size C~m:~

C~m~ = F~rw~ / F~aw~

= 1,218 / 10,020 (in)

= 0.122 in \< 0.38 in (designated single groove weld)

If adding the value of the outside intermittent fillet weld (0.31"), the
weld

size even can go further smaller size.

So, the design of the weld for the upstream end flange is satisfactory.

Per the above calculations and analysis, the downstream end flange and
the bellow flange have met the codes of ASME VIII, Div.1.

**5.2. [Upstream End Flange, Slide Flange and Bellow
Flange]{.underline}**

References: ME-439317-2, MD-439240

Endcap (module 2-3), #0988348/0, DESY -- MKS1

and Figure 5.2

![](media/image8.png){width="5.749305555555556in"
height="5.049305555555556in"}

Figure 5.2. [Upstream flange with the below flange]{.underline}

Since there some information are missing at this moment, so it is to
make the best assumption as followings:

1\. The upstream end flange connect to the downstream stream end flange
of the

bellow (see drawing Endcap (module 2-3), #0988348/0, DESY -- MKS1).

> 2\. The above two flanges are connected by a slide flange.
>
> 3\. A = 50.35", G = 47.889", T~1~ = 1.57", T~2~ = 1.57", B~1~ =
> 47.34", B~2~ = 42.24",
>
> C = 49.21"

4\. n = 24, D = 0.551"

E = 28 x 10^6^ psi, Modules of elasticity of the flange materials

Material: SA 182-F304L stainless steel, F~a~ = S~f~ = 16,700 psi at the
operating

temp., -20^0^ F to 100^0^ F (per Table 1A, part D, Section II, ASME
Boiler &

Pressure Vessel Code 1995 edition).

S~b~ = 25 ksi for bolt material: SA-193, S304 for operating

temperature, -20^0^ F to 100^0^ F (per Table 3, part D, Section II, ASME

Boiler & Pressure Vessel Code 1995 edition).

Per section Y-5.1, Y-5.2, figure Y-5.1.2 of Appendix Y,

Section 2-4, figure 2-4 of Appendix 2 of ASME VIII:

B~2~ \> ½ C,

So it is defined as: Class 2, Category 3 flange

In order to reduce the amount of work required as shown on section 5.1,
per section Y-9, of Appendix Y, ASME VIII, Div.1, we apply the method of
estimating flange thickness and bolting for the downstream flanges.

t~a~ = 2.45 \[Mp / ((πC --nD)S~f~)\]^1/2^

(per eq. 39 of Y-9, appendix Y, ASME VIII, Div.1)

where:

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~ (per sectionY-3, Appendix Y, ASME
VIII)

H~D~ = 0.785 B~1~^2^ P

= 0.785 x 47.34^2^ in^2^ x 14.7 psi

= 25,861 lbs.

H = 0.785G^2^P

= 0.785 x 47.889^2^ in^2^ x 14.7 psi

= 26,484 lbs.

H~T~ = H - H~D~

= (26,464 -- 25,861) lbs.

= 603 lbs.

h~G~ = (C -- G)/2

= (49.21 -- 47.889)in / 2

= 0.6605 in

h~D~ = (C -- B) / 2

= ( 49.21 -- 47.34) / 2

= 0.935 in

h~T~ = (h~D~ + h~G~) / 2

= 0.7978 in

H~G~ = W -- H, gasket load

(per section 2-3, appendix 2, ASME VIII)

W = H + H~p,~

flange design load for the operating condition

(per section 2-5(c)(1), appendix 2, ASME VIII)

which leads to:

H~G~ = W -- H

= H + H~p~ -- H

= H~p~

= 2b x 3.14GmP

= 2 x 0.281 in x 3.14 x 47.889 in x 3.50 x 14.7 psi

= 4,348 lbs.

M~p~ = H~D~h~D~ + H~T~h~T~ + H~G~h~G~

= (25,861 x 0.935 + 603 x 0.7978 + 4,348 x 0.6605) lbs-in

= (24,180 + 481 + 2,872) lbs-in

= 27,533 lbs-in

t~a~ = 2.45 \[Mp / ((πC --nD)S~f~)\]^1/2^

= 2.45 x \[ 27,533 lbs-in / ((π 49.21 in -- 24 x 0.551 in) x 16,700
psi)\]^1/2^

= 2.45 x 0.1078 in

= 0.264 in

t~b~ = 0.56 B~1~ (P/S~f~)^1/2^

= 0.56 x 47.34 in x (14.7 psi / 16,700 psi)^1/2^

= 0.56 x 47.34 in x 0.0297

= 0.788 in

Since t~c~ is the greater of t~a~ or t~b~, so:

t~c~ = t~b~ = 0.788"

Per equation 42 of A-7, Appendix Y, ASME VIII, Div. 1:

A'~b~ = (H + 2 M~p~ / (A -- C)) ÷ S~b~

= (26,464 lbs + 2 x 27,533 lbs-in / (50.35 -- 49.21) in ) ÷ 25,000 psi

= (26,464 lbs + 48,304 lbs) / 25,000 psi

= 2.9907 in^2^

Per Table Y-9.1, Appendix Y, ASME VIII, Div.1, it is found:

t~1~ = 1.1t~a~ = 0.29" \< T~1~ = 1.57"

t~2~ = 1.1t~c~ = 0.867" \< T~2~ = 1.57"

So the thickness of the flanges are ok for the upstream end flanges.

*[Welds Analysis and Calculations:]{.underline}*

The weld configurations are the same as the upstream flange, it designed
per section UW-13 (e)(2), Fig. UW-13.2, section 2-3 and Fig. 2-4(4a),
Appendix 2 of ASME VIII, Div.1

Also, since the valus of L~sd~ for upstream end flange is the same as
for the downstream end flange), "I" value for upstream end flange is
similar as the downstream flange, all these lead to a similar values of
F~rw~ and C~m~ for the upstream end and downstream end flanges, so the
same sizes and types of the weld design of the upstream end flange are
satisfactory.

Per the above calculations, the upstream end flange and the bellow
flange have met the codes of ASME VIII, Div.1.

**5.3. [Flanges #8, #10, #2 and #3]{.underline}**

In order to reduce the amount of work required as shown on section 5.1,
per section Y-9, of Appendix Y, ASME VIII, Div.1, we apply the method of
estimating flange thickness and bolting for the above flanges.

Since the information of the most mating flanges are not available at
this moment, I made some engineering judgments per the best of my logic
knowledge.

Table 5.3, [Estimated Calculated Flange thickness and bolting VS. the
designated data]{.underline}

  ---------------- -------- ------------ ------------ ------------ ------------
                            Flanges (see                           
                            drawing                                
                            ME-439317)                             

  nomenclature     unit     #8           #10          #2           #3

                            MC-439243    MC-439246    MC-439251    MC-439252

  P                psi      14.7         14.7         14.7         14.7

  Sf               psi      18,800       18,800       18,800       18,800

  A                in       10.945       5.512        25.2         25.2

  B1               in       8.386        3.334        21.65        21.65

  B2               in       3.75         3.334        0            0

  C                in       9.646        4.882        23.622       23.622

  G                in       8.793        4.012        22.636       22.636

  Classification            3            1            3            3

  Categorization            3            3            3            3

  n                         24           4            16           16

  D                in       0.3125       0.394        0.394        0.394

  m                         3.5          3.5          3.5          3.5

  b                         0.21         0.21         0.21         0.21

  HD               lbs      812          128          5409         5409

  H                lbs      892          186          5913         5913

  HT               lbs      81           57           504          504

  hG               in       0.4265       0.435        0.493        0.493

  hD               in       0.63         0.774        0.986        0.986

  hT               in       0.5283       0.6045       0.7395       0.7395

  HG               lbs      597          272          1536         1536

  MP               lbs-in   808          252          6463         6463

  ta               in       0.1064       0.0766       0.1744       0.1744

  tb               in       0.1313       0.0522       0.3390       0.3390

  tc               in       0.1313       0.0766       0.3390       0.3390

  A\'b             in\^2    0.1137       0.0525       0.7502       0.7502

  t1               in       0.1171       0.0766       0.1918       0.1918

  t2               in       0.1444       0.0766       0.3729       0.3729

  Ab               in\^2    0.1193       0.0525       0.7877       0.7877

  t1d              in       0.787        0.512        1.393        0.79

  t2d              in       N/A          0.512        N/A          N/A

  Abd              in\^2    1.088        0.271        1.085        1.085
  ---------------- -------- ------------ ------------ ------------ ------------

The most nomenclatures of the Table can be found from section 5.1 of
this note, also from the appendix 2 and appendix Y of ASME VIII, Div.1,
except:

t~1~, Estimated calculated flange 1 thickness

t~2~, Estimated calculated flange 2 thickness

A~b~, Estimated calculated root diameter area of the bolts.

t~1d,~ Designated flange 1 thickness

t~2d~, Dersignated flange 2 thickness

A~bd~, Designated root diameter area of the bolts.

Per table 5.3, it is found that all t~1~ \> t~1d~; t~2~ \> t~2d~ and
A~b~ \> A~bd~ for all respective flanges, so the flange design are
satisfactory.

Per figure UW-13.2, section UW-13, appendix 2 of ASME VIII, Div. 1 and
drawing ME-439317, it is found that all flange welds are designed per
the related codes.

6.  **[Finite Element Analysis (FEA) For the Vessel Weldment,
    Bottom]{.underline}**

> **[Supports, and Top Lifting Fixtures]{.underline}**

Reference drawing: ME-439317-1 (2)

Figure: i.1.

Per drawing ME-439317, it is found that there are 4 lifting points which
are symmetrical in both in xy and yz planes, it is also found that the 4
bottom supports only mirrored about y axis rotation (assuming that the
center of the vessel is the rotational point).

It is found that:

W~ws~ = 2,264 lbs. the weight of the vessel shell weldment

W~other~ ≈ 3,005 lbs.,

The weight of coldmass shields, support posts, support tubes weldment,

other attachments and mice. accessories.

W~tot~ ≈ W~ws~ + W~other~ = 5,269 lbs

Figures 6.1 & 6.2 are the calculated stresses from the FEA simulating
model, which is based on the 3.9 MHZ vessel weldment with the following
boundary conditions:

W~ws~ = 2,264 lbs applied to the model based the vessel physical
properties;

W~other~ ≈ 3,005 lbs external additional force applied to the model
through the 2 top

surfaces of coldmass port;

All (4) bottom saddle supports are assumed as simple support.

Figure 6.1, [Stress Values in X, Y & Z directions]{.underline}

![](media/image9.png){width="6.375in" height="3.654166666666667in"}

![](media/image10.png){width="6.159027777777778in"
height="8.752083333333333in"}

Figure 6.2, [The Von Mises Stress & the Deform.Display of the Vessel w/4
Bottom Supts.]{.underline}

![](media/image11.png){width="6.074305555555555in"
height="8.749305555555555in"}

Figure 6.3, [The Von Mises Stress & the Deform.Display of the Vessel w/4
Top Lift Pts.]{.underline}

![](media/image12.png){width="5.855555555555555in"
height="8.746527777777779in"}

Figure 6.4, [The Von Mises Stress & the Deform.Display of the Vessel
w/vacuum Cond.]{.underline}

It is found that the maximum Von Mises stress is about f~vm~ = 3.58 ksi
\< F~a~=15 ksi,

The higher stress areas are: bottom pad base support areas and port
opening area, however, all are far below the allowable stress F~a~.

Figure 6.3 is the Von Mises stress and deformation display of the vessel
weldment model, which is simulating the vessel under the same boundary
condition except to lift the vessel from 4 top pick points (instead of
supporting by 4 bottom locations). It is found that the maximum Vom
Mises stress is about 2.08 ksi, and the max. displacement is about
0.0024".

The simulating results of the vessel under the vacuum working
environment can be found from Figure 6.4, the following are the lists of
boundary condition:

1.  All open flanges and ports are connected with blind flanges with the
    matching

> thickness.

2.  All outside surfaces of the sealed vessel experienced pressure of
    14.7 psi.

> (Assuming all internal surfaces under pressure of zero)

3\. W~ws~ = 4,429 lbs. the weight of the vessel shell weldment and the
blind flanges.

W~other~ ≈ 3,000 lbs.,

The weight of coldmass shields, support posts, support tubes weldment,

other attachments and mice. accessories.

W~tot~ ≈ W~ws~ + W~other~ = 7,429 lbs

3.  All 4 bottom saddle supports are assumed as simple support.

It is found from Figure 6.4 that the maximum Von Mises is about f~vm~ =
7.18 ksi, and the most stress area is located around the top corner area
of the pad base plate of the bottom support which is next to the
upstream or downstream flange. The transition areas from the end flange
to the vessel shell, and the main coupler port opening areas are also
the relatively higher stress area, however, all calculated stress values
are less than the allowable stress F~a~ under the current boundary
condition. It is also found that the max. displacement (\~0.01") is
around the center of the end blind flange, that is due to the vacuum
pressure. All the results from FEA data confirm that the design of the
bottom supports, top lifting fixtures and the vessel weldment assembly
are satisfactory per the applicable codes.

7.  **[Appendix]{.underline}**

    1.  **[The Lists of the Design Drawings of 3.9 MHz, 3^rd^ harmonic
        Vacuum]{.underline}**

> **[Vessel (Fermilab)]{.underline}**
>
> 1\. Vacuum Vessel Weldment Assy. ME -- 439317 -- 1 (2)
>
> 2\. Vessel Shell ME -- 439262
>
> 3\. Port Fixed Coldmass MD -- 439251
>
> 4\. Port Sliding Coldmass MD -- 439252
>
> 5\. Pick Point Weldment MC -- 439257
>
> 6\. Pad Weld Upper MC -- 439242
>
> 7\. Vessel Support Weldment MC -- 439254
>
> 8\. Pad Base Support MC -- 439241
>
> 9\. Flange MC Support MD -- 439243

10\. Flange Vessel Electronics MD -- 439246

11\. Bracket, Weldment MC LH MC - 439238

12\. Bracket, Weldment MC LH MC -- 439239

13\. Flange, Cryostat End MD - 439240

14\. Rib, Stiffening MC -- 439253

15\. Gusset, Outrigger MC -- 439247

16\. Plate, Outrigger MC -- 439248