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Page 1 530.352 Materials Selection Lecture #11 Materials Selection Software Tuesday October 4 th, 2005.

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Presentation on theme: "Page 1 530.352 Materials Selection Lecture #11 Materials Selection Software Tuesday October 4 th, 2005."— Presentation transcript:

1 Page 1 530.352 Materials Selection Lecture #11 Materials Selection Software Tuesday October 4 th, 2005

2 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

3 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.

4 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

5 Page 5 Material indices: Performance = f [F,G,M] p = f [(F unctional requirements ), (G eometric constraints ), (M aterial properties )]

6 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?

7 Page 7 Function, object, constraint...  Function Tie Beam Shaft Column  Constraint Stiffness Strength Geometry Corrosion  Objective Minimum cost Minimum weight Maximum energy storage etc.

8 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 ).

9 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

10 Page 10 Table leg design:  Design goals minimize weight maximize slenderness  Constraint resistance to buckling

11 Page 11 Modeling a table leg:  Mass m =  r 2 l   Buckling load P crit =  2 EI =  3 Er 4 l 2 4l 2

12 Page 12 Minimizing weight :  Mass of legs: m = [4P /  ] 1/4 [l] 2 [  / E 1/2 ] M 1 = E 1/2 / 

13 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

14 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”

15 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)

16 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

17 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”

18 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 

19 Page 19 Stored energy / mass :  Energy / mass is : U/m = 1/4 R 2  2  Same for all materials ???

20 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]

21 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

22 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 = 


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