Materials Selection and Design: Introduction Outline Introduction Design Requirements Exampls: - Example 1: Strong and light Tie-Rod - Example 2: Stiff & ight Tension Members - - Example 4: ight and Strong Beam - Example 5: ight and Stiff Beam Materials Attributes: physical, mechanical, thermal, electrical, economic, environmental. Function Process Shape For selection, one must establish a link between materials and function, with shape and process playing also an important role (We will focus on the function part) Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/1 Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/2 Design Requirements Function: support a load, withstand temperature, transmit heat, etc. What does the component do? Objective: make thing cheaply, light weight, increase safety, etc., or combinations of these. What is to be maximized or minimized? Constraints: make thing cheaply, light weight, increase safety, etc., or combinations of these. What essential conditions to be met? Free variables: Which design variables are free? From which we obtain: - Screening Criteria: go / no-go criteria - Ranking Criteria: an ordering of the materials that go Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/3 A = x-area Function: Example 1: Strong and light Tie-Rod Tie-rod Tie-rod is common mechanical component. Tie-rod must carry tensile force, F, w/o failure. is usually fixed by design. While strong, need lightweight, or low mass. Objective: Minimize mass: M = A. (1) Constraints: = fixed length must not fail under load F (A can carry the load) F/A σ f. (2) Free variables: Material choice Section area A fixed Performance index: Eliminate A in (1) and (2): m = F σ f Chose materials with smallest /σ f Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/4
Example 1: Strong and light Tie-Rod Maximize P = σ f / P is performance index Example 1: Strong and light Tie-Rod Performance Index where a higher number gives better performance Consider log σ f vs log P = 1000 P = 100 P = 10 For fixed P, look at log σ f = log + C For fixed P, you look for ines of Slope = 1. Each ine of Slope = 1 have the same P values! But NOT the same materials properties (σ f or ) e.g. some less dense (lighter). For fixed P: log P = log σ f log = constant = C Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/5 Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/6 Example 2: Stiff & ight Tension Members Example 2: Stiff & ight Tension Members c c F, δ Bar must not lengthen by more than δ under force F; must have initial length. Stiffness relation: Mass of bar: F (σ = Eε) c 2 = E δ M =c 2 Eliminate the "free" design parameter, c: M = F2 δ specified by application E minimize for small M P = E for stiff tension members log P = log E log = C Maximize the Performance Index: P = E (stiff, light tension members) Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/7 Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/8
Example 3: Strong & ight Torsion Members Bar must carry a moment, Mt must have a length. Strength relation: Mass of bar: f M =πr 2 N = 2M t πr 3 N is a safety factor Eliminate the "free" design parameter, R: specified by application M = ( 2 π NM t ) 2/3 2/3 f minimize for small M Maximize the Performance Index: (strong, light torsion members) P = 2/3 f Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/9 Maximize Performance Material Index log = 3/2 [ log + log P ] ys P= Need to consider lines of constant P with slope of 3/2 on log-scale Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/10 2/3 Additional constraint E.g. require P 10 and f 300 MPa. All parallel lines have same performance index. 100 This identifies search region- shaded 300 MPa However, P=30 has 1/3 the mass of P=10 (mass 1/P ). 30 10 3 P = 10 All materials that lie on these lines will perform equally for strength-per-mass basis. However, each line has a different Materials M index, or overall Performance P index. Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/11 Strength vs Density Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/12
Example 4: ight and Stiff Beam F b b d=deflection Bending is common mode of loading in engineering, e.g., golf clubs, wing spars, floor joists. ight, square beam (A=b 2 ) with length, loaded in bending must meet a constraint on its stiffness, S, so that it does not deflect more than d with load F. Stiffness Constraint S= F d C 1 EI 3 Eliminating Area, A: m 12S C 1 =C 1 E b 4 3 = C 1 E A 2 12 3 12 1/2 ( 3 ) Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/13 Mass m = A. Constraint If beam remains square, the ight, Stiff Beam is one with largest P = E1/2 If only beam height can change (not A), then P = (E 1/3 /) (Car door) I b 3 w If only beam width can change (not A), then P = (E/) E 1/2 (note in first bracket) Maximize the Performance Index: P = σ 1/2 Strength, σf (MPa) Example 5: Strong & ight Bending Members 10 4 bending members Ceramics Cermets 10 3 PMCs Steels 10 2 grain Metal alloys Adapted from Fig. 6.22, 10 Polymers Callister 6e. grain 1 wood 0.1 0.1 1 10 30 Density, (Mg/m 3 ) slope = 2 Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/14 Examples of Materials Indices Strong & ight Tension vs. torsion Members Function, Objective, and Constraint Index Tie, minimum weight, stiffness E/ Beam, minimum weight, stiffness E 1/2 / Beam, minimum weight, strength σ 2/3 / Beam, minimum cost, stiffness E 1/2 /C m Beam, minimum cost, strength σ 2/3 /C m Column, minimum cost, buckling load E 1/2 /C m Spring, minimum weight for given energy storage σ YS2 /E Thermal insulation, minimum cost, heat flux 1/(α C m ) Electromagnet, maximum field, temperature rise κ C p C m =cost/mass α =thermal cond κ =elec. cond Strength, σf (MPa) 10 4 Ceramics 10 3 Cermets PMCs Steels 10 2 grain Metal alloys 10 Polymers grain 1 tension 0.1 members 0.1 1 10 30 Density, (Mg/m 3 ) wood slope = 1 slope = 3/2 torsion members Adapted from Fig. 6.22, Callister 6e. Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/15 Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/16
Next time: Continue Materials Selection Dr. M. Medraj Mech. Eng. Dept. - Concordia University Mech321 lecture 20/17