Narvik University College (Høgskolen i Narvik) EXAMINATION TASK COURSE STE6289 Modern Materials and Computations (Moderne materialer og beregninger 7.5 stp.) CLASS: Master students in Engineering Design (Ingeniørdesign (5ID)) DATE: Friday. December 205 TIME: 9.00-2.00, 3 hours NUMBER OF PAGES: 8 (inclusive Appendix: tables and formulae) UTILITIES: No printed or handwritten aids are allowed (e.g. dictionaries, cell phones, lap tops of any kind), except the paper copy of the report from the projects that are handed in at the exam. Certain, simple calculator is allowed (it should only have simple numerical and trigonometrical functions. The calculator should not have a graphical display, and if inspected it should be easy to identify). CONTACT PERSON DURING THE EXAM: professor Annette Meidell phone 76 96 62 25, mob. 995 60 739 or professor Dag Lukkassen tlf. 95 40 545. The exam can be answered in norwegian, scandinavian or english. Attach the final report concerning the project given in this course as the last subtask in this exam (mark it/them with your exam number not your name). Attach either one report in two paper copies or one paper copy and one electronically USB or CD (remember to mark it with student exam number only, and not your name). All papers must be handed in (except the Exam task itself). The total number of subtasks is 0, all having same weight. In Task "Project" the report(s) from the project should be attached and counts 2/5 of the total grade.
Figure 0.: A cantilever sandwich beam with an end point load. Task a) Describe the general basic properties of plastics. b) Explain how a structure with negative Poissons ratio will behave in contrast to a material or structure with positive Poissons ratio. Make a sketch and explain. c) How can you increase the stiffness and strength of a structure, when keeping the weight the same (or almost the same)? Task 2 In this task we are going to design a sandwich beam of length L = 3m such that the stiffness is P/δ = 23N/mm. It is required that the core has thickness t c = 00mm and depth b = 00mm and consists of a foam material with shear modulus G c = 30MPa, Poissons number ν c = 0 (i.e. E c = 2(+ν)30 = 2(+0)30 = 60MPa), see Figure 0.. The facings have Young s modulus of value E f = 20000MPa (Steel). We will use the approximation formula (EI) eq = D = E fbt f d 2. (0.) 2 a) Use the approximation d = t c in order to estimate the thickness of the facings t f such that the above requirements are met. b) How can we find the thickness of the facings t f if we use the exact identity d = t c + t f (you do not need to find the numerical value)? c) Is t f smaller or larger in the case a) than in the case b). Explain without calculating the value of t f. 2
Task 3 Consider a periodic symmetric unidirectional material structure which consists of two isotropic materials with a Y cell like the one in Figure 0.2 (note that this structure is not square symmetric). Figure 0.2: A periodic Y-cell. a) By using ANSYS, we find that the following two-dimensional boundary conditions (plane strain) u (0, x 2 ) = 0, u (2, x 2 ) = give the elastic energy W = 00. Use the equation for the energy W = 2 e C e Y, to find the element C of the effective stiffness matrix C given in (0.3). b) By using ANSYS, we find that the following two-dimensional boundary conditions (plane strain) u 2 (x, 0) = 0, u 2 (x, ) = give the elastic energy W = 200. Use the equation for the energy W = 2 e C e Y, 3
to find the element C 2222 of the effective stiffness matrix C given in (0.3). c) By using ANSYS, we find that the following two-dimensional boundary conditions (plane strain) u (0, x 2 ) = u 2 (x, 0) = 0, u (2, x 2 ) = u 2 (x, ) = /2 give the elastic energy W = 300. Use this and your answers in a) and b) to find the element C 22 (= C 22) of the effective stiffness matrix C given in (0.3). Task Projects Attach either one report in two paper copies or one paper copy and one electronically USB or CD (remember to mark it with your student exam number only, and not your name). 4
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, (simply supported) central load, P Three-point bend, (simply supported) 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 P L3 B δ (EI) s = P L eq B 2 (AG) eq M x = P L B 3 T x = P B 4 3 8 2 2 48 4 4 2 384 5 8 8 2 92 4 8 2 384 8 2 2 Deflection: Deflection: δ = δ b + δ s = δ = δ b + δ s = P L3 P L +. B (EI) eq B 2 (AG) eq P L3 P L +. B (EI) eq B 2 (AG) eq 5
Maximum face stress: Wrinkling stress: Flexural rigidity: (we can use if (EI) eq = E fbt 3 f 6 σ f = M x max t c E f 2 (EI) eq. σ cr = 0.5 3 E f E c G c + E fbt f d 2 2 (EI) eq = E fbt f d 2 2 + E cbt 3 c 2 (0.2) and The shear stiffness is given by ( ) 2 d 3 > 00 or t f 6E f t f d 2 E c t 3 c > 00). d t f > 5.77 and the shear stress (AG) eq = bd2 G c t c τ c = T x max. db 0.. 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 2 ( + ν) ( 2ν), G = E 2 ( + ν). 6
0.2. Locally and globally orthotropic structure: Stress/strain-relation: σ C C22 C33 0 0 0 σ 22 C22 C2222 C2233 0 0 0 σ 33 σ 2 = C33 C3322 C3333 0 0 0 0 0 0 C22 0 0 σ 23 0 0 0 0 C2323 0 σ 3 0 0 0 0 0 C33 0.3. Square symmetric unidirectional two-phase structure: Stress/strain-relation: σ σ 22 σ 33 σ 2 σ 23 σ 3 = K + G T K G T l 0 0 0 K G T K + G T l 0 0 0 l l n 0 0 0 0 0 0 G T,45 0 0 0 0 0 0 G L 0 0 0 0 0 0 G L ɛ ɛ 22 ɛ 33 γ 2 γ 23 γ 3 e e 22 e 33 γ 2 γ 23 γ 3, (0.3). Effective compliance matrix: Relations: E T ν T E T ν L E L ν T ET ET ν L E L ν L EL ν L E L E L 0 0 0 G T,45 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 G L 0 0 0 0 0 G L 0. G T = E T 2 ( + ν T ), 4 E T = G T + K + 4 (ν L )2, EL 7
l = ν L2K, n = E L + 4 (ν L) 2 K, EL = p o E o + p I E I + 4(ν o ν I 2 ( ) 2 (p o + p I ), (0.4) 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 ), (0.5) K I K o K I K K i = E i 2 ( + ν i ) ( 2ν i ) and G i = E i 2 ( + ν i ) ). 8