1. Page 1 530.352 Materials Selection Lecture #10: Materials Selection Charts Monday October 3, 2005

2. Page 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. Page 3 Range of material properties :

4. Page 4 Simple solution :  List in rank order and choose cut-off point. Steel210 GPa Glass 80 GPa Al 69 GPa Wood 8 GPa

5. Page 5 However...  Often must consider more that one property. I.e. space structures must be light and stiff !!!  Can take ratios : specific modulus E /  = … specific strength  failure /  = …

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

7. Page 7 Plate deflection :  = 0.67 Mga 2  E t 3 Mass =  a 2 t  solve for t and substitute: M = (0.67g /  ) 1/2  a 4 (  / E 1/3 ) 3/2 M 1 = (  / E 1/3 ) 2a t  W=mg

8. Page 8 Beam deflections : Square beam of length L :  = 4 L 3 F / E t 4 M = L t 2  Solve and substitute: M = (4L 5 F /  ) 1/2 (  / E 1/2 ) M 2 = (  / E 1/2 )  F L

9. Page 9 Beam buckling : ? Post of length L : m =  r 2 l  P crit =  2 EI =  3 Er 4 l 2 4l 2 Solve and substitute: M = (4L 5 F /  ) 1/2 (  2 / E) 1/2 M 2 = (  / E 1/2 )

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

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

12. Page 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. Page 13 Material grouping into classes : Fig. 4.2 in Materials Selection (p 34) Density (g / cm 3 ) 0.1 1.0 10 Modulus (E) (GPa) 1000 10 0.1 alloys ceramics composites woods polymers elastomers foams

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

15. Page 15 Contour plots : Fig. 4.2 in Materials Selection (p 34) Density (g / cm 3 ) 0.1 1.0 10 Modulus (E) (GPa) 1000 10 0.1 alloys ceramics composites woods polymers elastomersfoams 10 4 m/s 10 3 m/s

16. Page 16 Adding contours to the plots : From beam buckling we get : E 1/2 /  = const. take log and rearranging gives: log (E) = 2 log (  ) + 2 log (const.) slope of log-log plot is: d log (E) / d log (  ) = 2 intercept of log-log plot is: related to const.

17. Page 17 Adding contours to the plots : From plate deflection we get : E 1/3 /  = const. take log and rearranging gives: log (E) = 3 log (  ) + 3 log (const.) slope of log-log plot is: d log (E) / d log (  ) = 3 intercept of log-log plot is: related to const.

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

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