Page 2 Material Selection - the basics: All materials Screening: apply property limits / eliminate those who cannot do the job Ranking: apply material indices / find best candidates Subset of materials Supporting info: Handbooks, software, WWW, etc. Prime candidates Local conditions: in-house expertise or equipment Final Material Choice
Page 3 Deriving property limits: Simple limits on material properties can be used to eliminate possible materials e.g. T operating = 250 o C Electrically insulating must be available in wire form etc.
Page 4 Deriving material indices: Combination of material properties Used when component characteristics can be achieved in more than one way: e.g. high stiffness high modulus increasing the cross-section changing the shape
Page 5 Material indices: Performance = f [F,G,M] p = f [(F unctional requirements ), (G eometric constraints ), (M aterial properties )]
Page 6 Function, objective, constraint: Function: what does component do? Objective: what is to be maximized -or- minimized? Constraints: what non-negotiable conditions must be met? what other conditions are desired?
Page 7 Function, object, constraint... Function Tie Beam Shaft Column Constraint Stiffness Strength Geometry Corrosion Objective Minimum cost Minimum weight Maximum energy storage etc.
Page 8 Procedure for deriving material indices: Define design requirements Develop an equation for the objective in terms of functional requirements, geometry and material properties. Identify the free (unspecified) variables. Develop constraint equations. Substitute for the free variables. Group the variables into three groups and determine: p = f 1 (F),f 2 (G),f 3 (M) Identify the Material Index (M 1 ).
Page 9 Table legs: Goal: light weight coffee table of daring simplicity: a flat sheet of glass with slender light weight legs. Legs must: be solid be light as possible support a load P without buckling
Page 10 Table leg design: Design goals minimize weight maximize slenderness Constraint resistance to buckling
Page 11 Modeling a table leg: Mass m = r 2 l Buckling load P crit = 2 EI = 3 Er 4 l 2 4l 2
Page 12 Minimizing weight : Mass of legs: m = [4P / ] 1/4 [l] 2 [ / E 1/2 ] M 1 = E 1/2 /
Page 13 Criterion for slenderness: Minimum leg radius P crit = 3 Er 4 4l 2 r = [4P / 3 ] 1/4 [l] 1/2 [1 / E ] 1/4 M 2 = E
Page 14 CES Software: CES software available in the HITS Computing Lab (Krieger 160) or Senior Design Computer Lab. Access it the following way: 1. Click “Start” menu 2. Go to “Programs” ->”Engineering Applications” ->“CES” -> “CES Selector”
Page 15 Table leg materials: Good : light weight: –woods ; composites ; ceramics slender (stiff) –CFRP ; ceramics Not good : polymers (too compliant) ; metals (too heavy - except Be)
Page 16 Table leg materials = 1/2 ; M 2 = E Make Modulus-density chart MaterialsM 1 M 2 Comment wood5-8 4-20cheap, reliable steel1.8 210poor M 1 CFRP4-830-200very good, expensive Ceramics4-8 150-1000 excellent but brittle
Page 17 Materials for Flywheels : Flywheels store energy Current flywheels are made out of : children’s toys –lead steam engines –cast iron modern electric vehicles –HSLA steels and composites Efficiency measured in “stored energy per unit weight”
Page 18 Stored energy : For a disc of radius (R) and thickness (t) rotating with angular velocity ( ), the energy (U) stored in the flywheel is : U = 1/2 J 2 = 1/4 R 4 t 2 The mass of the disk is : m= R 2 t
Page 19 Stored energy / mass : Energy / mass is : U/m = 1/4 R 2 2 Same for all materials ???
Page 20 Centrifugal stress : Maximum principal stress in a spinning disk of uniform thickness : max = [(3+ )/8] R 2 2 This sets the upper limit of ; U/m = [2/(3+ )] [ f / ] M = f / kJ / kg]
Page 21 Materials for flywheels : MaterialM [kJ/kg]Comments Ceramics200-2,000Brittle in tension. CFRP200-500best performance good choice. GFRP100-400cheaper than CFRP excellent choice. Steel, Al, Ti, Mg100-200Steel cheapest Cast iron8-10high density Lead alloys3high density
Page 22 Why use lead and cast iron ?? Children’s toys use these -- why ?? Cannot accelerate to the burst velocity If angular velocity is limited by the drive mechanism (pull string) then : –U = 1/4 R 4 t 2 –M 2 =