hw7sol

# hw7sol - ME 363 1(a ρ c = ρ solid i0.7 = 4000 × 0.7 =...

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Unformatted text preview: ME 363 1. (a) ρ c = ρ solid i0.7 = 4000 × 0.7 = 2800 kg / m 3 ρc 2800 ρ sinter = = = 3, 266 kg / m 3 3 3 ⎛ ΔL ⎞ ⎜1 − ⎟ L⎠ ⎝ (1 − 0.05) (b) 4000 - 3266 = 0.184 4000 UTS = 180e-4×0.184 = 86.22 MPa P= E = E0 (1-1.9 P + .9 P 2 ) = 300 (1-1.9 × 0.184 + .9 × 0.1842 ) = 204.3 GPa Homework #7 2. L Pavg = ∫ P ⋅ dx = 1 L∫ 0 L L = L ( L ) (1 − x L )dx 0.1 x ) ( 2 0.1 x ∫ L − x L2 dx L0 L 3 0.1 ⎡ x 2 ⎤ −x 3L2 ⎥ 0 ⎣ ⎦ L ⎢ 2L 0.1 ⎡ L 0.1 = −L ⎤= = 0.017 3⎦ 6 L⎣ 2 = ∴1.7% k = k0 (1 − P ) kavg ( )( ) = k0 ⎡1 − 0.1 x 1− x ⎤ L L⎦ ⎣ = k0 (1 − Pavg ) = 0.983k0 = 0.786 W / mK 3. (a) x = 0.4 Ec = x ⋅ E f + (1 − x ) Em = 0.4 ⋅ 400 + (1 − 0.4)100 = 220 GPa Pf Pm = Af E f Am Em = .4 × 400 160 = = 2.67 .6 × 100 60 Pc = Pf + Pm = Pf + Pf 2.67 = 1.375Pf ∴ Pf = .73Pc (b) Ec = x ⋅ 275 + (1 − x )100 = 220 GPa x = 0.685 ∴ 68.5%. 4. (a) Ld - L = 0.07 Ld L = (1- 0.07 ) Ld Ld = 30 = 32.26 mm 0.93 L0 = (1 + 0.08) Ld = (1.08)( 32.26 ) = 35 mm (b) ρ f = (1 − 0.09 ) ρ = 0.91ρ Because the linear shrinkage during firing is 8%, ρ d = (1 − 0.08) ρ f 3 = 0.78ρ f = 0.71ρ ...
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