Download Problem Set #5 - Materials Selection | EMA 4714 and more Assignments Materials science in PDF only on Docsity! Problem Set 5 EMA 4714 - Materials Selection and Failure Analysis Due Wednesday, February 28, 2001 - 11:45 am Dieter Exercise 8.8 In an aerospace application, total cost of a component can be expressed by: Ct = Cf + CmW + PW where: Cf = cost of fabrication Cm = material coast, $/lb W = weight of component, lb P = penalty factor by which performance is jeopardized, $/lb [a] Discuss the various strategies available to minimize Ct. [b] Determine whether material B can be economically substituted for material A if [1] there is no weight penalty factor and [2] the weight penallty is $100 per lb. *************************************************************************** [a] [i] Lower Cf by changing to a lower cost process [or a place where processing costs are lower] or increase production rate [reducing cost per component]. [ii] Lower material cost [Cm]. [iii] Lower component weight [W] by redesign - involves analysis of function and property relationship. [iv] Lower W by selection of alternate [higher strength] material which is likely to be more expensive. If CB > CA, but WB < WA, selection of material B will be justified when the total costs of using B are less than those for using A: CBWB < CAWA CB < CA[WA/WB] CB < CA/[1-f] where f = [WA - WB]/WA [b] assuming Cf(A) = Cf(B), when weight is penalized [P factor introduced], use of B is justified if: CBWB + PWB < CAWA + PWA CB < CA/[1-f] + Pf/[1-f] f = [90 - 2]/90 = 0.978 2100B 901A W, lbCm, $/lbmaterial [1] with P = 0, use of B is justified when CB < 1/[1-0.978] < $45.45 [2] with P =100, use of B is justified when CB < 1/0.022 + [100][0.978]/0.022 = $4490 Dieter Exercise 8.11 Select a group of candidate materials for an energy-storing flywheel using the Ashby charts. First develop the performance index M from the following information about flywheels: Consider the flywheel to be a solid disk of radius R and thickness t, rotating with an angular velocity ω. The energy stored in the flywheel is: U = Jω2/2 = [1/2][πρR4t/2]ω2 J = polar moment of inertia for the disk The quantity to be maximized [objective function] is the kinetic energy per unit mass. The maximum stress [constraint] in the spinning disk is given by: σmax = [(3 + ν)/8]ρR2ω2 *************************************************************************** flywheel mass, m, = πR2tρ Kinetic energy stored, U, = [π/4]ρR4tω2 Objective function, U/m = R2ω2/4; R2 = 4[U/m]/ω2 Constraint, σmax < σt: σ = {(3 + ν)/8}ρR2ω2; R2 = σt/{[(3 + ν)/8][ρω2} Clearing free variable, R, U/m = [2/(3 + ν)][σt/ρ], where M = σt/ρ
