Preliminary Calculations of CPA Properties of G10 Ultimate stress - - PDF document

Preliminary Calculations of CPA

Properties of G10 ≔ σultimate 32000 psi psi Ultimate stress ≔ ρG10 1800 ―― kg kg m3 Density ≔ EG10 19.1 GPa GPa Modulus ≔ αG10 ⋅ ⋅ 9.6 10−6 ――― cm cm ⋅ cm cm K Coefficient of thermal expansion Geometric Properties ≔ LpD 6 m Length of proto-DUNE CPA ≔ LD 12 m Length of DUNE CPA ≔ width 1.15 m Width of CPA ≔ thick 0.25 in in Thickness of CPA plane ≔ a 1 in in Width of CPA edge stiffener ≔ b 3 in in Height of CPA edge stiffener ≔ Temp 83 K Liquid Argon Temperature Calculate Shrinkage ≔ ∆length = ⋅ ⋅ LpD Temp αG10 4.78 mm mm Change in proto-DUNE CPA length ≔ ∆width = ⋅ ⋅ width Temp αG10 0.92 mm mm Change in CPA width Compute properties of CPA cross section ≔ Ix = ⋅ 2 ⎛ ⎜ ⎝ ―― ⋅ a b3 12 ⎞ ⎟ ⎠ 4.5 in in4 A 2 ( b) 6 i

2

≔ Areabars = ⋅ 2 ( ( ⋅ a b) ) 6 in in2 ≔ Areasheet = ⋅ width thick 11.32 in in2 Calculate Weight of CPA ≔ qG10 = ⋅ ⋅ ρG10 ⎛ ⎝ + Areabars Areasheet⎞ ⎠ g 1.13 ―― lbf lbf in in ≔ Wbars = ⋅ ⋅ ⋅ ρG10 Areabars LpD g 92.17 lbf lbf ≔ Wsheet = ⋅ ⋅ ⋅ ρG10 Areasheet LpD g 173.87 lbf lbf ≔ WCPA = + Wbars Wsheet 266.04 lbf lbf Calculate deformation and stress due to lifting CPA with gravity loading -- CPA simply supported at ends ≔ Mmax = ―――― ⋅ qG10 LpD

2

8 7855.49 ⋅ lbf lbf in Maximum moment in CPA ≔ σmax = ――― ⋅ Mmax ― b 2 Ix 2618.5 ps psi Maximum stress in CPA Maximum deflection of CPA ≔ ∆max = ――――― ⋅ ⋅ 5 qG10 LpD

4

⋅ ⋅ 384 EG10 Ix 3.66 in in Calculate Deformation Due to Differential Pressure in DUNE ≔ q1 = ⋅ ( ( + 1.27 Pa Pa 0.84 Pa Pa) ) width 0.01 ―― lbf lbf in in ≔ q2 = ⋅ 0.84 Pa Pa width 0.01 ―― lbf lbf in in

≔ y1( (x) ) ⋅ ―――― q1 ⋅ ⋅ 6 EG10 Ix ⎛ ⎜ ⎜ ⎝ − + − ――― ⋅ LD x3 6 ――― x5 ⋅ 20 LD ――― ⋅ LD

3 x

20 ――― ⋅ LD

3 x

6 ⎞ ⎟ ⎟ ⎠ ≔ y2( (x) ) ⋅ ―――― q2 ⋅ ⋅ 2 EG10 Ix ⎛ ⎜ ⎝ − − ――― ⋅ LD x3 6 ― x4 12 ――― ⋅ LD

3 x

12 ⎞ ⎟ ⎠ ≔ y( (x) ) − y1( (x) ) y2( (x) ) Lateral deformation of CPA due to differential pressure = y ⎛ ⎜ ⎝ ―― LD 2 ⎞ ⎟ ⎠ −0.07 in in ≔ x , ‥ 0 in in ⋅ .1 in in LD

• 0.06
• 0.05
• 0.05
• 0.04
• 0.03
• 0.02
• 0.02
• 0.01
• 0.08
• 0.07

0.01 90 135 180 225 270 315 360 405 450 45 495

x ( (in in) ) y( (x) ) ( (in in) ) ≔ M( (x) ) − + ⋅ ――― ⋅ q1 LD 6 x ⋅ ―― q1 ⋅ 6 LD x3 ⋅ ― q2 2 ⎛ ⎝ − ⋅ LD x x2 ⎞ ⎠ Calculate Net force and moment due to pressure differential on CPA ≔ Fnet = − ――― ⋅ q1 LD 2 ⋅ q2 LD 0.67 lbf lbf

≔ Znet = ―――――――― + ⋅ ――― ⋅ −q1 LD 2 ―― LD 3 ――― ⋅ q2 LD

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2 Fnet 3.81 m Distance from top of CPA to point where the next forces is acting ≔ ∆CPAcg = ――― ⋅ Znet Fnet WCPA 9.56 mm mm Distance CPA center of gravity will move laterally to counteract the moment due to the differntial pressure ≔ WAPA 900 lbf lbf Weight of APA ≔ ∆CPAcg = ―――――― ⋅ Znet Fnet + WCPA ⋅ 2 ―― WAPA 2 2.18 mm mm Distance CPA center of gravity will move laterally to counteract the moment due to the pressure differential when tied to the APAs which help prevent rotation. Calculate Weight Carried by CPA Support ≔ WFC = ⋅ 200 kg kg g 440.92 lbf lbf Weight of FC - half of this weight is supported by a CPA ≔ Fstrap = + WCPA ⋅ 4 ⎛ ⎜ ⎝ ―― WFC 2 ⎞ ⎟ ⎠ 1147.89 lbf lbf ≔ SF 10 Safety factor ≔ Astrap = ―――― ⋅ SF Fstrap σultimate 0.36 in in2 Cross sectional area required = ‾‾‾‾‾ Astrap 0.6 in in Size of strap if it was square. Calculate the Moment and Lateral Motion Due to FC Hanging from CPA ≔ e 4 in in Distance from CPA centerline to center of gravity of FC L 2 8

Length of FC ≔ LFC 2.8 m = WFC 440.92 lbf lbf ≔ Xcg = ―――――― ⋅ ⋅ 2 e ―― WFC 2 + ⋅ 2 ―― WFC 2 WCPA 2.49 in in ≔ Zcg = ――――――――――――――― + + ⋅ ―― WFC 2 ―― LFC 2 ⋅ ―― WFC 2 ⎛ ⎜ ⎝ − LpD ―― LFC 2 ⎞ ⎟ ⎠ ⋅ WCPA ―― LpD 2 + ⋅ 2 ―― WFC 2 WCPA 3 m