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530.352 Materials Selection

530.352 Materials Selection. Lecture #10: Materials Selection Charts Monday October 3, 2005. Common wisdom :. Material properties limit performance . Performance can be maximized by comparing and selecting the appropriate material. Always consider a wide range of materials.

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530.352 Materials Selection

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  1. 530.352 Materials Selection Lecture #10: Materials Selection ChartsMonday October 3, 2005

  2. Common wisdom : • Material properties limit performance. • Performance can be maximized by comparing and selecting the appropriate material. • Always consider a wide range of materials.

  3. Range of material properties :

  4. Simple solution : • List in rank order and choose cut-off point.Steel 210 GPa Glass 80 GPa Al 69 GPa Wood 8 GPa

  5. However ... • Often must consider more that one property.I.e. space structures must be light and stiff !!! • Can take ratios : • specific modulus E / r = … • specific strength sfailure / r = …

  6. Ratios of properties : Relative importance depends on design criteria !!!

  7. 2 a  t W = m g Plate deflection :  = 0.67 Mga2  E t3 Mass = a2 t  solve for t and substitute: M = (0.67g / )1/2a4 (/ E1/3)3/2 M1 = (/ E1/3)

  8. L  F Beam deflections : Square beam of length L :  = 4 L3 F / E t4 M = L t2  Solve and substitute: M = (4L5F / )1/2 ( / E 1/2) M2 = ( / E 1/2)

  9. Beam buckling : • Post of length L : • m = p r2 l r • Pcrit = p2 EI = p3Er 4 l2 4l2 • Solve and substitute: • M = (4L5F / )1/2 (2 / E)1/2 • M2 = ( / E 1/2) ? ?

  10. The Ashby solution : Materials Selection Charts hi,lo hi,hi Property #1 lo,lo lo,hi Property #2

  11. The Ashby solution : Materials Selection Charts hi,lo subrange Property #1 lo,lo lo,hi Property #2

  12. Examples of MS charts : • Materials Selection , Ch. 4 • Modulus - Density (p 37) • Strength - Density (p 39) • Modulus - Strength (p 42) • Loss coefficient - Modulus (p 48) • Conductivity - Diffusivity (p 49) • Expansion - Modulus (p 52) • Modulus - Cost (p 57) • etc.

  13. Material grouping into classes : Fig. 4.2 in Materials Selection (p 34) 1000 10 0.1 ceramics alloys composites Modulus (E) (GPa) woods polymers foams elastomers 0.1 1.0 10 Density (g / cm3)

  14. Adding contours to the plots : From wave equation we know : vel = (E / r )1/2 take log and rearranging gives: log (E) = log (r) + 2 log (vel) slope of log-log plot is: d log (E) / d log (r) = 1 (for const. vel) intercept of log-log plot is: related to vel.

  15. Contour plots : Fig. 4.2 in Materials Selection (p 34) 1000 10 0.1 ceramics alloys 104 m/s composites Modulus (E) (GPa) woods polymers 103 m/s foams elastomers 0.1 1.0 10 Density (g / cm3)

  16. Adding contours to the plots : From beam buckling we get : E1/2 / r = const. take log and rearranging gives: log (E) = 2 log (r) + 2 log (const.) slope of log-log plot is: d log (E) / d log (r) = 2 intercept of log-log plot is: related to const.

  17. Adding contours to the plots : From plate deflection we get : E1/3 / r = const. take log and rearranging gives: log (E) = 3 log (r) + 3 log (const.) slope of log-log plot is: d log (E) / d log (r) = 3 intercept of log-log plot is: related to const.

  18. Contours : * see Fig. 4.3 in Materials Selection (p 37)

  19. Can use both to compare : B A m Property #1 Property #2

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