Download presentation

Presentation is loading. Please wait.

Published byCarly Lanman Modified over 2 years ago

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

Similar presentations

Presentation is loading. Please wait....

OK

Adding Up In Chunks.

Adding Up In Chunks.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on depreciation of indian rupee 2013 Ppt on eisenmenger syndrome vsd Ppt on formal education meaning Ppt on 60 years of indian parliament pictures Ppt on unity in diversity indonesia Ppt on holographic technology in education Cell surface display ppt on tv Ppt on overview of financial services in india Ppt on tens and ones for grade 1 Ppt on verbs for grade 5