HØGSKOLEN I NARVIK - SIVILINGENIØRUTDANNINGEN EKSAMEN OG LØSNINGSFORSLAG I KURSET STE 689 MODERNE MATERIALER OG BEREG- NINGER KLASSE: 5ID, DATO:. Desember 9 TID: 9.-., 3 timer EMNESTØRRELSE: 7.5 stp. ANTALL SIDER: (inklusiv Appendix: tabell og formler) TILLATTE HJELPEMIDLER: Bestemt, enkel kalkulator tillatt, noe som betyr at kalkulatoren kun skal inneha enkle numeriske og trigonometriske funksjoner. Kalkulatoren skal ikke ha grafisk display, og ved kontroll skal den være enkelt identifiserbar. Ingen trykte eller håndskrevne hjelpemidler tillatt. KONTAKTPERSONER UNDER EKSAMEN: professor Annette Meidell tlf. 995 6 739 eller professor Dag Lukkassen tlf. 95 4 545. Eksamensbesvarelsen kan gis på norsk, skandinavisk eller engelsk. Totalt antall deloppgaver er 9, og de teller like mye. I Task Project er det de 3 prosjektene i kurset som skal legges ved.
Task a) What is a biopolymer? Answer: According to the American Society for Testing and Materials (ASTM), biopolymers are degradable polymers in which degradation results from the action of naturally occurring micro-organisms such as bacteria, fungi and algae. Biopolymers are an alternative to petroleum-based polymers (traditional plastics) produced by living organisms. The term biopolymers is often used for all polymers that are made from natural renewable resources and/or are completely biodegradable. Biopolymers can be produced by biological systems like microorganisms, plants and animals, or chemically synthesized from biological starting materials (e.g. sugars, starch, natural fats or oils, etc.). Most of the biopolymers are biodegradable, compostable or will biodegrade in landfill (time can vary from a couple of days to even years, but they will eventually degrade). Additional answer: Biopolymers can be used in all kinds of applications, and with all kinds of production techniques, like injection moulding, thermoforming and blow moulding! Some biopolymers can directly replace synthetic plastics in traditional applications, while others possess unique properties that may open new applications. b) Explain how a structure with negative Poissons ratio will behave in contrast to a material or structure with positive Poissons ratio. Answer: Instead of contracting in the direction perpendicular to the loading plane if the structure is exposed to tension, it will expand (the opposite reaction compared to a rubber band which gets thinner in the middle when stretching it). In an isotropic material structure (a material which does not have a preferred orientation) the allowable range of Poisson s ratio is from. to +.5. Normally, materials have a Poissons ratio <ν<.5. Metals have a Poissons numer of about.3. Hierarchical structures and foams have also been made with negative Poissons ratio (also called auxetic materials, anti-rubber, ) ν <, and Poisson s ratio >, which give exciting respond to loads. In stead of contracting in the direction perpenidcular to the loading plane if it is exposed to tension, it will expand (the opposite reaction compared to a rubber band which gets thinner in the middle when stretching it), see an example of a structure with negative Poissons number
Figure.: An example of a structure with negative Poissons ratio. in Figure.. c) What is the main ideas with reiterated homogenization? Answer: For a reiterated structure (hierarchical structure), see an example in Figure.. we can use the homogenization theory to obtain for instance the effective thermal conductivity (for instance if the structure is a reiterated chessboard structure) for each level of the structure, beginning at the inner level of hierarchy. By using the homogenized (effective moduli) as the next input value for the next level, we can continue in this way until there are no more levels. The final value is the effective conductivity for the structure. This technique is a very efficient method for finding the effective moduli in complicated structures which is difficult to analyze by using ordinary finite elemet programs due to the different size in the areas. This makes the solution difficult to obtain due to the large number of elements required to obtain the solution. Task a) Find the deflection of the sandwich beam with ends built in and that has a midpoint load P N, see Figure.3 by using the approximation D E fbt f d. The length of the sandwich L 5mm, thickness of the facings t f is mm and t c is 4mm, b mm. The facings are aluminium plates with Youngs 3
Figure.: A hierarchical structure. modulus of E f MPa (GRP), strength σ f,cr MPa. The core material is a PVC material with shear modulus G c 4MPa (8kg/m 3 PVC foam), (E c MPa), shear strength τ c,cr.5mpa. Answer: The total deflection is described as PL 3 δ δ b + δ s + PL, 9 4(AG) eq where the bending contribution is δ b PL 3 P L 3 P (5) 3 ³ ³ 9 (EI) Ef bt eq 9 f d 9 (4+) 3.87 9 4 (P ) (mm/n) 3.87 9 4 () 3.87 9mm, and the shear contribution is δ s PL PL P 5 ³ ³ 7.436 4 (P ) (mm/n) 4(AG) eq bd 4 G c t c 4 (4+) 4 4 7.436 4 () (mm/n) 7.436mm. 4
Figure.3: A sandwich beam with ends built in and a midpoint load. The total deflection is δ δ b + δ s 3.87 9 + 7.436.39mm. b) Is the approximation for valid? If not find the correct deflection. Answer: Therequirementsduetotheapproximationsisvalidif µ d 3 3 t f (or d t f 4 µ 4 543. > 4> 5.77) then the first term of.3 is less than % of the second, and can therefore be neglected, and since 6E f t f d E c t 3 c 6 (4) (4) 3 3.59, which is less than,hence, the third term of.3 is not less than % of the second, and can therefore not be neglected. The assumption D E fbt f d 5
is not valid! We have to use D E fbt f d + E cbt 3 c Ã! (4 + ) µ 4 3 +.68 9 +5.333 3 7. 734 3 9 which gives that the bending contribution δ b PL 3 9 P L 3 9 ³ P (5) 3 9 (. 734 3 9 ) 3.753 9 4 (P ) (mm/n) 3.753 9 4 () 3.753 9mm, and the shear contribution δ s PL PL P 5 ³ ³ 7.436 4 (P ) (mm/n) 4(AG) eq bd 4 G c t c 4 (4+) 4 4 7.436 4 () (mm/n) 7.436mm. The total correct deflection is δ δ b + δ s 3.753 9 + 7.436.9mm. c) Calculate the face thickness, t f, for maximum allowable stress. Is the thickness given in a) satisfactory? Answer:The stress in the facings σ f M x maxt c E PL f tc E 8 f ³ (EI) Ef bt eq f d 5 8 4 ³ 56.5, () tf 4 t f 6 4 5 ³ 8 () tf 4
here d t c, thus the face thickness for maximum allowable stress (Or: t f 56.5 56.5.565mm. σ f σ f M x maxt c E f M x maxt c E 5 f 8 4. 5 9 σ f µ µ Ef bt f d () tf 4. 5 9 (.5 9 ) t f () 4.565 ). Thethicknessgivenina)isnotsatisfactory. Itshouldbeatleast.565mm in order to avoid that the allowable face stress is exceeded. d) Calculate the maximum shear stress in the core. Is it satisfactory? Answer: The maximum shear stress in the core τ c T x max db P db P db.9 5MPa 4 () which is satisfactory since τ c,cr.5 MPa.. e) What is the wrinkling stress? Will it wrinkle? Answer: The wrinkling stress (can be found from (.)) σ cr.5 3p E f E c G c.5 3 4 5 MPa. The face will not wrinkle since σ cr >σ f MPa. 7
Task 3 Assume that you have a periodic square symmetric unidirectional structure with a periodic cell Y [, ] [, ]. The periodic material structure consists of two isotropic materials with Young s modulus E o GPa and E I GPa, Poisson s ratio ν o and ν I.4 and volume fractions p I p o / (where index o and I refer to the white and black material, respectively). Describe as detailed as possible how you proceed in order to compute the value of the effective bulk modulus K. Answer: For the computation we can solve the two-dimensional cell problem by using plane strain. The effective stress/strain-relation is hσ i hσ i hσ 33 i hσ i hσ 3 i hσ 3 i K + G T K G T l K G T K + G T l l l n G T,45 G L G L he i he i he 33 i hγ i hγ 3 i hγ 3 i (see the compendium). In order to compute the effective in-plane moduli K we solve the cell problem for hei [,,,,,, ] T and compute the corresponding values K hσ i /. Alternatively we can compute the strain energy W and compute K from the identity W hei C hei Y (.) Since Y 4, we obtain from (.) that the strain energy W corresponding to the average strain vector hei [,,,,,, ] T is K + G T K G T l W K G T K + G T l l l n G T,45 4 G L G L 8
(K + G T )+(K G T ) (K G T )+(K + G T ) 4 K K 4 W (K +K )4 (4K )4K 8 which gives us that K W/8. We can use uniform boundary conditions on each face of the square structure. More precisely, for hei [,,,,,, ] T the boundary conditions are: u (,x ), u (,x ), u (x, ), u (x, ). Projects Attach the copies of the three projects in this course (legg ved kopiene av de tre prosjektene dere har gjort i dette kurset). 9
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 B 3 B 4 PL3 B δ (EI) s PL eq B (AG) eq M x PL B 3 T x P B 4 3 8 48 4 4 384 5 8 8 9 4 8 384 8 Deflection: Maximum face stress: δ δ b + δ s PL3 PL +. B B (AG) eq Wrinkling stress: σ f M x max t c E f. σ cr.5 3p E f E c G c (.)
Flexural rigidity: (we can use if E fbt 3 f 6 + E fbt f d E fbt f d + E cbt 3 c (.3) and The shear stiffness is given by µ d 3 > or t f 6E f t f d E c t 3 c > ). d t f > 5.77 and the shear stress (AG) eq bd G c t c τ c T x max. db.. Isotropic materials 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 (+ν)( ν), G E (+ν)... Square symmetric unidirectional two-phase structure: Stress/strain-relation:
hσ i hσ i hσ 33 i hσ i hσ 3 i hσ 3 i K + G T K G T l K G T K + G T l l l n G T,45 G L G L he i he i he 33 i hγ i hγ 3 i hγ 3 i. Effectivecompliancematrix: Relations: E T ν T E T ν L E L ν T ET E T ν L E L ν L E L ν L EL E L G T,45 G L G L. 4 E T G T G T E T (+ν T ), + K + 4(ν L ), EL l ν LK,n EL +4(ν L) K, EL p o E o + p I E I + 4(ν o ν I ) ³ (p o + p I ), (.4) 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 ), (.5) K I K o K I K K i E i (+ν i )( ν i ) and G i E i (+ν i ) ).