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  ![](media/image1.png){width="0.875in"   **Fermilab**
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**Particle Physics Division**

**Mechanical Department Engineering Note**

# Number: MD-ENG- 117 Date: September 18, 2006

Project Internal Reference:

Project: SciBooNE

Title: The Brief Structural Analysis for the Main Frame of the Scibar &

EC Detector

Author(s): Edward Chi

Reviewer(s): Ang Lee

Key Words: Frame, Structure, Tubing, Allowable Stress, Von Mises Stress,
Welds,

Shear and Bending Stress.

Abstract Summary: The main frame is specially designed for supporting

the 35 ton force which is composed of the 64 scibar

module layers, 64 Electron Catcher modules and

numerous accessorial components of the detector. The

several working stresses of the main frame were calculated by

the simulated Finite Element Model, so as to calculate the

deflection of the frame under the applying load. The

allowable stresses, welds have been extensively discussed and

calculated per the related applicable specifications and codes.

Applicable Codes:

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

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

"Design of Weldments", By Omer W. Blodgett

1\. *[Overview]{.underline}*

As showing in Figure 1.1, the support frame is for housing the Scibar,
Electron Catcher (EC) detectors, and other numerous accessorial devices
and instruments, and the total weight of the system (live and dead) is
about 35 tons (70,000 lbs).

![](media/image2.png){width="7.235416666666667in"
height="4.286111111111111in"}

Fig. 1.1, [The Isometric View of the Scibar and EC Detector of the
SciBoone]{.underline}

For the simplicity, it is reasonable to assume that the structure of the
main frame has to be designed to support the 35 tons applying load. The
following page has more descriptions about the details of the applying
load and the distributions.

The most critical load case for the main frame is that the main frame is
lifted up for moving through four lifting pads at the top of the frame
while all the components of the detector are installed in place. There
is an extensive discussion regarding such loading case in section 3.

![](media/image3.png){width="3.5791666666666666in"
height="4.252083333333333in"}

Fig. 1.2, [The Support Frame of Scibar & EC Detectors]{.underline}.

(2D drawing: ME-435903)

The Figure 1.2 is an iso. View of the frame and the mechanical drawing
ME-435903 has detail specifications for the construction of the frame.

The 64 Scibar module layers are support by frame through 64 mounting
brackets, the horizontal EC (with 32 modules) and the vertical EC (w/32
modules) are installed in the downstream of the support frame, figure
1.3 is showing the inside iso installation view of the support system
without the dark room around.

The support frame is designed to be built by the steel structural
tubing, the specifications of the steel tubing are:

8" x 8" x ½" structural steel tubing,

ASTM A500, Grade B

F~y~ = 46 ksi

A = 14.4 in^2^,

I~xx~ = I~yy~ = 131 in^4^

S~xx~ = S~yy~ = 32.9 in^3^

Ґ = 3.03 in

![](media/image4.png){width="3.3618055555555557in"
height="3.9972222222222222in"}

Fig. 1.3, [Inside installation iso. view of the Scibar & EC detectors
w/support frame]{.underline}

2\. *[Discussions and Calculations of the Allowable
Stresses]{.underline}*

Considering the complex of the force distribution of the support frame,
I built a FEA model to simulate the work load case of the support frame
to calculate the working stress and displacement under different
boundary conditions. However, it is still necessary to find out the
allowable working stresses F~a~, F~b~ & F~v~ under the current working
boundary condition.

Assuming the square tube L~x~=140 in (from fig. 1.2) under the loading
case of "simple beam with uniformly distributed load" (#1 case of page
2-296, ASD 9^th^ edition):

Per section B5.1d of ASD 9^th^ edition,

b = 8" -- 3x0.5"= 6.50 in

d = 8"

t~f~ = 0.5" = t~w~

Per table B5.1 of section B5:

190/(F~y~)^1/2^ = 28

238/(F~y~)^1/2^ = 35

b/t = 6.50/0.5 = 13 \< 190/(F~y~)^1/2^ = 28 \< 238/(F~y~)^1/2^ = 35

Per equation F1-2 of Sections F1.1 and eq. F1-5 of section F1.2:

L~c~ = (76b~f~)/(F~y~)^1/2^

= (76 x 8 in) / (46ksi)^1/2^

= 89 in \< L~x~=140 in

So:

F~b~ = 0.6 F~y~

= [27.6 ksi]{.underline}

Per section F4 of ASD 9^th^ edition:

380 / /(F~y~)^1/2^ = 56

h = 8" -- 2 x 0.5"

= 7.0 in

h/t~w~ = 7 in/0.5 in

=14 \< 380/(F~y~)^1/2^ = 56

Per eq F4-1 of section F4 of ASD 9^th^ edition:

F~v~ = 0.40 F~y~

= [18.4 ksi]{.underline}

The allowable stress for the vertical square column tube L~y~=126 in:

Per section E2 of ASD 9^th^ edition:

Assuming K = 1.0 for the loading case of the support frame, then:

C~c~ = (2π^2^E / F~y~ )^1/2^

= 111

KL / ґ = 1.0 x 126 in / 3.03 in

= 41.58 \< C~c~ = 111

The allowable stress is:

F~a~ = \[1- (KL/ґ)^2^ / 2 C~c~^2^\]F~y~ ÷ \[5/3 + 3 (KL/ґ)/8C~c~ -
(KL/ґ)^3^/8C~c~^3^\]

= \[1- (41.58)^2^ / 2x111^2^\]x 46 ksi ÷ \[5/3 + 3x 41.58/8x111 -
(41.58)^3^/8x 111^3^

= 42.77 ksi ÷ (1.667 + 0.14 -- 0.00657)

= [23.75 ksi]{.underline}

3\. *[Discussios and Analysis of the FEM results:]{.underline}*

As mentioned in section 2, several FEA models were built to simulating
the main frame force loading conditions, and calculate the working
stresses and displacements of the support frame under simulating
boundary conditions of the frame. Fig. 3.1 is showing the model
simulating the boundary conditions:

Due to the capacity of the computer, the model is assumed the boundary
condition is symmetrical about the YZ plane @ where x=0.0, so it was use
half of the frame.

The mass weight of the frame W~f~ = 10,500 lbs (50% of it) is
distributed through the frame model.

The forces from EC vertical modules, EC horizontal modules, 64 Scibar
module layers, vertical and top horizontal PMT assemblies, dark-house (6
sides with top side also acting as rated service floor as shown in
fig. 1) and other accessorial devices are distributed stimulatingly per
the design and application of the whole detector support system.

![](media/image5.png){width="4.6506944444444445in"
height="7.542361111111111in"}

Figure 3.1, [The Boundary Conditions for the Subjected FEA
Model]{.underline}

(The main frame is lifted with the full detector components are loaded)

![](media/image6.png){width="5.59375in" height="8.252083333333333in"}

Figure 3.2, [The Calculated Stress Distribution When the Detector is
Lifted]{.underline}

![](media/image7.png){width="6.6097222222222225in"
height="4.240972222222222in"}

Table 3.1, [The Summary of the reaction force for the FEA
model]{.underline} .

![](media/image8.png){width="6.663888888888889in"
height="3.984027777777778in"}

Table 3.2, [The Summary of the calculated stresses for the FEA
model]{.underline}

It is found from Figure 3.2 that the max. Von Mises stress **F~vm~ =
7.75 ksi** which is less than the allowable stress **F~a~ = 23.75 ksi**.
(see section2, page 5).

Table 3.2 summarized all max. & min. stresses in different directions
from the simulated model. As you can see, all the values of the stress
are less than the allowable stresses F~v~ and F~b~ respectively.

The deflections:

The SciBooNE Experiment set the criteria of the deflection in Y
direction (the direction of the gravitation) is +/- 0.059" (+/- 1.5 mm).

From the Figure 3.2, it is found that the maximum deflection of the
frame is about 0.0312", which is less than the experimental defined
value of 0.059".

4\. *[The discussions of the welds:]{.underline}*

All welding metals are E70, where F~u~ = 70 ksi

Then, the allowable stress of the welding metal: F~wv~ = 0.30 x 70 ksi =
21ksi

(per Table J2.5, Chapter J, ASM, 9^th^ edition)

The actual loading case for the tubing sections of the frame (ref. to
fig. 1.2 of page 3, fig.1.3 of page 4 and fig. 3 of page 6) is between
the "simple beam to the fixed beam at both ends, force uniformly
distributed." (See cases #1 & #15 of section 2, ASD, 9^th^ edition).

For the simple and conservative sake, we chose the beam (tube) connected
as: **fixed at both ends** **with uniformly distributed loads w**. We
also picked the longest beam support span L = 140" as the case we'll
discuss (see fig. 1.2 in page 3). See fig. 4.1 for the force
distribution.

Fig. 4.1, [Force distribution of the frame tubing beam]{.underline}.

Conservatively assuming P = 17.5 tons = 35,000 lbs applying to the
subjected beam along,

Where:

L = 140.0",

w = P/L = 35,000 lbs / 140" = 250 lbs/in

R~1~ = R~2~ = P/2 =17,500 lbs.

M~max~ = wL^2^ / 12 = (250 x 140^2^) / 12 = 408,334 lbs-in

Figures 4.2 is partial view from the engineering drawing and figure 4.3
is the configuration of the tubing welds in cross view.

![](media/image9.png){width="4.000694444444444in"
height="2.777083333333333in"}

Fig. 4.2, The partial view of the main frame drawing ME-435903 (related
to the tubing

Weld in discussed area)

![](media/image10.png){width="2.751388888888889in" height="2.44375in"}

Figure 4.2. The weld configuration of the connect tubing beam to treat
as lines

*Find the geometric properties of the welds* (treated as lines):

From figure 4,2, it is found that:

Length L = 2 (b+d) = 32 in where b = 8 in = d

I~z~ = I~y~ = d^2^ (3b + d) ÷ 6

= 341.33 in^3^

S~z~ = Sy = d (3b + d) ÷ 3

= 85.33 in^2^

Compute the components of the force in the welds:

The primary shear stress due to direct load f~v~

f~v~ = R1 ÷ weld length

=17,500 lbs ÷ 32.0 in

= 547 lbs/in ↓

The bending stress f~b~ about the Z axis:

f~b~ = M~max~ / I~z~

= (408,334 ÷ 85.33) lbs/in

= 4,786 lbs/in

The resultant force f~t~

f~t~ = (f~v~^2^ + f~b~^2^) ^1/2^

= (547^2^ + 4,786^2^)^1/2^ lbs/in

= 4,817 lbs/in

Find the fillet weld size C, per drawing; all are fillet welds with E70
weld metal: Allowable stress of the weld metal F~wv~:

F~wv~ = 0.30 x 70 ksi = 21ksi, for E70 weld metal

The allowable weld size C:

C = f~t~ / (0.707 x F~wv~ )

= 4,817 lbs/in ÷ (0.707 x 21,000) lbs/in^2^

= 0.3245 in

Per drawing ME-435903 (also see fig. 4.2), it was found that the weld
sizes are 0.50", so the design is satisfactory.

5\. *[The conclusions:]{.underline}*

The calculated working stresses are less than the allowable stress of
the support frame; the calculated deflection values are less than the
experimental designated deflection values; the designed welding sizes
are much larger than the calculated required sizes. Up to those ends,
the design of the main frame is satisfactory to the applying force
distributions of the Scibar and Electron Catcher detector.

6\. *[Some direct related reference drawings:]{.underline}*

ME -- 435903 -1

ME -- 435903 -2

ME -- 444071 -1

ME -- 444071 -2

In order to view more details of the drawings, please go to the link:

<http://home.fnal.gov/~edchi/SciBoone/EC_Scibar/drawings/>

The site contains the master drawing list and the pdf format type of 2d
drawings