HØGSKOLEN I NARVIK - SIVILINGENIØRUTDANNINGEN

Transkript

1 HØGSKOLEN I NARVIK - SIVILINGENIØRUTDANNINGEN EKSAMEN I FAGET STE 6243 MODERNE MATERIALER KLASSE: 5ID DATO: 7 Oktober 2005 TID: , 3 timer ANTALL SIDER: 7 (inklusiv Appendix: tabell og formler) TILLATTE HJELPEMIDLER: Kalkulator/regnemaskin med tekstlagringsmulighet er ikke tillatt Ingen trykte eller håndskrevne hjelpemidler tillatt KONTAKTPERSONER UNDER EKSAMEN: Annette Meidell tlf og/eller Dag Lukkassen tlf Eksamensbesvarelsen kan gis på norsk, skandinavisk og/eller engelsk Totalt antall deloppgaver som teller like mye: 8! I tillegg skal det innleveres kopier av de 2 prosjektene i kurset

2 Task a) What is the difference between a thermosetting plastic and a thermoplastic? b) What is a metamaterial? Give an example of such a material, and how it behaves c) What is Layer Manufacturing Technology (LMT)? Task 2 a) What is a failure-mode map? Explain how it works and what it can be used for b) For a given sandwich-problem the following failure-mode transition equations are given: The Face yield (mode ) - Face wrinkling (mode 2) transition equation is given by log ρ c =log The Face yield (mode )- Core shear (mode 3) transition equation is given by log ρ c =log log t f L The Face wrinkling (mode 2) - Core shear (mode 3) transition equation is given by log ρ c =log log t f L Illustrate these transition equations as graphs and find the corresponding failure mode-map (explain the procedure you use) c) What is the approximate total midpoint deflection of a simply supported sandwich beam (see Figure 0) with a central load P =000N, shear modulus in the core G c =20MPa (foam),young s modulus in the faces and core E f = MPa, length L = 000 mm, face thickness t f =2mm, core thickness t c =50mm and width b =00mm? (you may assume that E c << E f ) 2

3 Figure 0: A simply supported (three-point bend) sandwich beam d) We recall that the effective conductivity λ eff of a regular chessboard structure (sjakkbrettstruktur), consisting of two isotropic materials with conductivities λ and λ 2, is given by the formulae λ eff = p λ λ 2 Use this formula and reiterated homogenization to find an approximate value of the effective conductivity of a so-called iterated chessboard structure illustrated as the body Ω is Figure 02 Here, the conductivity of the white material is α =6 and the conductivity of the black material is α 2 =8 Figure 02: Iterated chessboard structure 3

4 e) Assume that we have an isotropic material with Young s modulus E and Poisson s ratio ν Find E and ν expressed by the shear modulus G and the bulk modulus K Projects Attach the copies of the two projects in this course (legg ved kopiene av de to prosjektene dere har gjort i dette kurset) 4

5 Appendix: Tables and formulae Mode of loading, (all beams of length L) Cantilever, end load, P Cantilever, Uniformly distributed load, q = P/L Three-point bend, central load, P Three-point bend, Uniformly distributed load, q = P/L Ends built in, Central load, P Ends built in, Uniformly distributed load, q = P/L δ b = B B 2 B 3 B 4 PL3 B δ (EI) s = PL eq B 2 (AG) eq M x = PL B 3 T x = P B Deflection: Maximum face stress: δ = δ b + δ s = PL3 PL + B (EI) eq B 2 (AG) eq Wrinkling stress: σ f = M x max t c E f 2(EI) eq σ cr =05 3p E f E c G c 5

6 Flexural rigidity: (we can use if (EI) eq = E fbt 3 f 6 + E fbt f d 2 2 (EI) eq = E fbt f d E cbt 3 c 2 and The shear stiffness is given by µ 2 d 3 > 00 or t f 6E f t f d 2 E c t 3 c > 00) (AG) eq = bd2 G c t c The maximum shear stress in the core is given by 0 Isotropic materials τ c = T x max db d t f > 577 For an isotropic material the shear modulus G and bulk modulus K in plane elasticity (plane strain) are related to the well known Young s modulus E and Poisson s ratio ν as follows: K = E 2(+ν)( 2ν), G = E 2(+ν) 6

7 02 Square symmetric unidirectional two-phase structure: Stress/strain-relation: hσ i hσ 22 i hσ 33 i hσ 2 i hσ 23 i hσ 3 i = K + G T K G T l K G T K + G T l l l n G T, G L G L he i he 22 i he 33 i hγ 2 i hγ 23 i hγ 3 i Effectivecompliancematrix: Relations: ν T ν T ET EL G T, G L 0 0 G L 0 G T = 2(+ν T ) 4 = + ET G T K + 4(ν L) 2 EL l = ν L2K,n = EL +4(ν L) 2 K EL = p o E o + p I E I + 4(ν o ν I ) 2 ³ 2 (p o + p I ), (0) K o K o K I K K I ν L = p o ν o + p I ν I ν o ν I K o (p o + p I ) (02) K I K o K I K K i = E i 2(+ν i )( 2ν i ) and G i = E i 2(+ν i ) ) 7