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

Preliminary Calculations of CPA Properties of G10 Ultimate stress ≔ 32000 psi σ ultimate psi kg kg Density ≔ 1800 ―― ρ G10 m 3 Modulus ≔ 19.1 GPa E G10 GPa 9.6 10 −6 ――― cm cm Coefficient of thermal expansion ≔ ⋅ ⋅ α G10 ⋅ cm cm K Geometric Properties Length of proto-DUNE CPA ≔ 6 m L pD Length of DUNE CPA ≔ 12 m L D Width of CPA ≔ 1.15 m width Thickness of CPA plane ≔ 0.25 in thick in Width of CPA edge stiffener ≔ 1 in a in Height of CPA edge stiffener ≔ 3 in b in Liquid Argon Temperature ≔ 83 K Temp Calculate Shrinkage Change in proto-DUNE CPA length ≔ ⋅ ⋅ = 4.78 mm ∆ length L pD Temp α G10 mm Change in CPA width ≔ ⋅ ⋅ = 0.92 mm ∆ width width Temp α G10 mm Compute properties of CPA cross section ⎛ ⎞ a b 3 ⋅ in 4 ≔ 2 ⋅ = 4.5 in ⎜ ―― ⎟ Ix 12 ⎝ ⎠ 2 ( b ) 2 6 i A

2 ( a b ) in 2 ≔ ⋅ ( ⋅ = 6 in ) Area bars in 2 ≔ ⋅ = 11.32 in Area sheet width thick Calculate Weight of CPA lbf lbf ρ G10 ⎛ Area sheet ⎞ ≔ ⋅ + ⋅ = 1.13 ―― q G10 Area bars ⎠ g ⎝ in in ≔ ⋅ ⋅ ⋅ = 92.17 lbf W bars ρ G10 Area bars L pD g lbf ≔ ⋅ ⋅ ⋅ = 173.87 lbf W sheet ρ G10 Area sheet L pD g lbf ≔ + = 266.04 lbf W CPA W bars W sheet lbf Calculate deformation and stress due to lifting CPA with gravity loading -- CPA simply supported at ends 2 ⋅ q G10 L pD Maximum moment in CPA ≔ ―――― = 7855.49 ⋅ M max lbf lbf in 8 b ⋅ M max ― 2 Maximum stress in CPA ≔ ――― = 2618.5 ps σ max psi Ix 4 5 q G10 L pD ⋅ ⋅ Maximum deflection of CPA ≔ = 3.66 in ――――― ∆ max in 384 E G10 Ix ⋅ ⋅ Calculate Deformation Due to Differential Pressure in DUNE lbf lbf ( Pa ) ≔ 1.27 Pa + 0.84 Pa ⋅ = 0.01 ―― q 1 ( Pa ) width in in lbf lbf ≔ 0.84 Pa ⋅ = 0.01 ―― q 2 Pa width in in

⎛ 3 x 3 x ⎞ L D x 3 ⋅ x 5 ⋅ ⋅ q 1 L D L D y 1 ( ( x ) ⎜ ⎟ ≔ ―――― ⋅ ――― − ――― + ――― − ――― ) 6 E G10 Ix ⋅ ⋅ 6 20 L D ⋅ 20 6 ⎜ ⎟ ⎝ ⎠ ⎛ 3 x ⎞ L D x 3 ⋅ x 4 ⋅ q 2 L D y 2 ( ( x ) ⎜ ⎟ ≔ ―――― ⋅ ――― − ― − ――― ) 2 E G10 Ix ⋅ ⋅ 6 12 12 ⎝ ⎠ y ( ( x ) y 1 ( ( x ) y 2 ( ( x ) Lateral deformation of CPA ≔ − ) ) ) due to differential pressure ⎛ ⎞ L D ―― = −0.07 in ⎜ ⎟ y in 2 ⎝ ⎠ ≔ 0 in , .1 in ⋅ ‥ x in in L D 0.01 0 0 45 90 135 180 225 270 315 360 405 450 495 -0.01 -0.02 -0.02 -0.03 y ( ( x ) ) ( in ) ( in ) -0.04 -0.05 -0.05 -0.06 -0.07 -0.08 x ( in ) ( in ) ⋅ q 1 L D q 1 q 2 x 2 ⎞ ⎛ M ( ( x ) x 3 ≔ ⋅ + ⋅ − ⋅ ⋅ − ) ――― ―― ― x L D x ⎝ ⎠ 6 6 L D ⋅ 2 Calculate Net force and moment due to pressure differential on CPA ⋅ q 1 L D ≔ ――― − ⋅ = 0.67 lbf F net q 2 L D lbf 2

2 − q 1 L D ⋅ ⋅ L D q 2 L D ――― ⋅ ―― + ――― 2 3 2 Distance from top of CPA to point ≔ ―――――――― = 3.81 m Z net F net where the next forces is acting ⋅ Z net F net Distance CPA center of gravity will ≔ ――― = 9.56 mm ∆ CPAcg mm W CPA move laterally to counteract the moment due to the differntial pressure Weight of APA ≔ 900 lbf W APA lbf ⋅ Z net F net Distance CPA center of ≔ = 2.18 mm ―――――― ∆ CPAcg mm W APA gravity will move laterally to + 2 ―― ⋅ W CPA counteract the moment due 2 to the pressure differential when tied to the APAs which help prevent rotation. Calculate Weight Carried by CPA Support Weight of FC - half of this weight is ≔ 200 kg ⋅ = 440.92 lbf W FC kg g lbf supported by a CPA ⎛ ⎞ W FC ≔ + 4 ⋅ ―― = 1147.89 lbf ⎜ ⎟ F strap W CPA lbf 2 ⎝ ⎠ Safety factor ≔ 10 SF ⋅ SF F strap in 2 Cross sectional area required ≔ = 0.36 in ―――― A strap σ ultimate ‾‾‾‾‾ Size of strap if it was square. = 0.6 in A strap in Calculate the Moment and Lateral Motion Due to FC Hanging from CPA Distance from CPA centerline ≔ 4 in e in to center of gravity of FC 2 8 L

Length of FC ≔ 2.8 m L FC = 440.92 lbf W FC lbf W FC 2 e ―― ⋅ ⋅ 2 ≔ ―――――― = 2.49 in X cg in W FC 2 ―― ⋅ + W CPA 2 ⎛ ⎞ W FC L FC W FC L FC L pD ―― ⋅ ―― + ―― ⋅ − ―― + W CPA ―― ⋅ ⎜ ⎟ L pD 2 2 2 2 2 ⎝ ⎠ ≔ ――――――――――――――― = 3 m Z cg W FC 2 ―― ⋅ + W CPA 2