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Solution (1) The deflection constraint requires that δ < δ max with δ max = 0. 01in : PR 2 PR 2 δ = ≤ δ Max ⇒ t ≥ (1) 2Et 2Eδ Max For a spherical cap, the mass m depends on the thickness and the surface area: m = 2π R 2 ( 1− cosθ ) t ρ (2) where R is the radius of curvature and θ is the included angle. Substituting for t and rearranging in terms of Functional, Geometric, and Material requirements gives the mass m1 of the shield required to meet the deflection constraint: P m R 4 ρ 1 ≥ π ( 1− cosθ ) (3) δ Max E This suggests mass is minimized by maximizing E/ρ , shown as one axis of the figure opposite. (2) The buckling criterion can be re-stated as: t 2 P S f × P ≤ PCrit = 0. 17E ⇒ t ≥ (4) R 2 0. 11 E (with Sf = 1.5). The mass is still given by equation (2). Thus the mass m2 required to resist buckling is: m 2πR 2 2 = ( 1− cosθ ) tρ Substituting for t and rearranging gives: 2π m P R 3 ρ 2 ≥ ( 1− cosθ ) (6) 0. 11 E This suggests mass is minimized by maximizing E 1/2 /ρ , shown as the other axis of the figure. (5) The figure is best used by employing the coupling line construction described in Reference 1, Section 9.2, pp. 241 – 245. Setting m 1 = m2 gives the equation for the coupling line: E 2 δ E = max ⇒ ρ 0. 11 R P ρ E = ρ 0. 68 (using the data given in the question, and remembering that the units of stress and pressure are 10 6 psi). Slide the box up the line until a small subset of materials is left. Coupling line Selection box © Granta Design, March 2009 44 E ρ
(3) The strength criterion can be restated as: S f PR 3 PR S f × σ = ≤ σ Limit ⇒ t ≥ (7) 2t 4σ Limit Substituting for t into the mass equation (2) above gives the mass m3 required to give sufficient strength: 3 3 ρ m3 ≥ π P R ( 1− cosθ ) (8) 2 σ Limit The mass m3 is minimized by maximizing σ lim it / ρ . It appears as one axis of the figure opposite. (4) The damage criterion can be restated as: S f PR 3 PR πa K I = σ πa = πa ≤ K IC ⇒ t ≥ (9) 2t 4K IC Substituting for t into the mass equation (2) above and rearranging: 3 3 / 2 3 ρ m4 ≥ π P a R ( 1− cosθ ) (10) 2 K IC This suggests mass is minimized by maximizing ΚΙC / ρ, shown as the other axis of the figure. The figure is used in the same way as the previous one. Setting m 3 = m4 gives the equation for the coupling line: K1c ρ σ limit = π a ⇒ ρ K1c ρ = 0. 4 σ limit ρ using the limiting crack size of 0.05 in. This line is plotted on the figure. A selection box with its corner on the line, as shown, correctly balances the constraint on strength and that on damage tolerance. Sliding this up the line, and exploring the intersection between the two selection stages gives the materials that best meet the four criteria. Coupling line Selection box © Granta Design, March 2009 45
Page 1: Cambridge University M A T E R I A
Page 4 and 5: © Granta Design, March 2009 2
Page 6 and 7: 1.3 Find the record for the Borosil
Page 8 and 9: 2. SEARCHING Searching lets you pul
Page 10 and 11: 3. SELECTING materials and process
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Page 16 and 17: Notes for instructors The procedure
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Page 54: Young's modulus, E (GPa) 1000 100 1

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