hw7sol

# hw7sol - ME 363 1. (a) Homework #7 ρ c = ρ solid i0.75 =...

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Unformatted text preview: ME 363 1. (a) Homework #7 ρ c = ρ solid i0.75 = 4000 × 0.75 = 3000 kg / m 3 ρc 3000 ρ sinter = = = 3,390 kg / m 3 3 3 ⎛ ΔL ⎞ ⎜1 − ⎟ L⎠ ⎝ (1 − 0.04 ) (b) 4000 - 3390 = 0.15 4000 UTS = 140e-4×0.15 = 76.83 MPa P= E = E0 (1-1.9 P + .9 P = 228 GPa 2 ) = 310 (1-1.9 × 0.15 + .9 × 0.152 ) 2. (a) x = 0.4 Ec = x ⋅ E f + (1 − x ) Em = 0.4 ⋅ 73 + (1 − 0.4)100 = 89.2 GPa Pf Pm = Af E f Am Em = .4 × 73 29.2 = = .487 .6 × 100 60 Pf .487 Pc = Pf + Pm = Pf + ∴ Pf = .33Pc (b) = 3.053Pf Ec = x ⋅ 380 + (1 − x )100 = 89.2 GPa x = −0.0385 ∴ cannot the exact Ec , so set x to 0. 3. (a) L = (1- 0.07 ) Ld Ld = 30 = 32.26 mm 0.93 Ld - L = 0.07 Ld 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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