This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EMSE 201 — Introduction to Materials Science & Engineering 8 November 2006 Name: SOLUTION Department of Materials Science and Engineering 1 of 6 Case Western Reserve University © 2006 Mark R. De Guire, except figures from Callister. Test #2 — 75 minutes; 150 points; 6 questions; 6 pages; 15% of course grade Partial credit will be given for correct setups and reasoning. Give units on numerical answers where appropriate. Please write all answers on these pages; use the backs if needed. 1) A rod, 25.0 cm long and with circular cross section, must be able to support a load of 44.4 kN in uniaxial tension along its length. For each of the four candidate materials listed below, a) (8 points) compute the minimum crosssectional area of the member to ensure that it will not yield (or in the case of aluminum oxide, not break) under the specified load; b) (10 points) rank the weight of the resulting members from 1 (lightest) to 4 (heaviest). Show your work in the space below. material yield strength, MPa density, g cm –3 crosssectional area, cm 2 rank by weight (1 lowest, 4 highest) annealed 1040 steel 355 7.85 1.25 4 (245 g) ascast 356.0 aluminum 124 2.69 3.58 3 (241 g) aluminum oxide, 96% 337* 3.72 1.32 1 (122 g) nylon 6,6 55 1.14 8.07 2 (230 g) *) fracture strength For a uniaxial force F on a specimen of uniform initial crosssectional area A , the engineering stress σ = F / A ( 2 pts ). To satisfy the criterion of no plastic deformation, we require that the stress not exceed the yield strength σ y (where we use the yield strength as an approximation for the stress at which yielding will begin) ( 2 pts ). Setting the stress equal to the yield strength, A = 44.4 kN/ σ y ( 1 pt for each correct numerical answer). The mass M of the rod will be given by AL ρ ( 3 pts ), where L is the length (specified as 25.0 cm) and ρ is the mass density. Substituting A = F / σ y gives M = FL ρ / σ y ( 3 pts ). The masses are listed in parentheses in the rightmost column of the table above. It was not necessary to compute the masses of the rods, as only a ranking was requested. Note that the only parameter that will vary with material in the expression for M is the ratio ρ / σ y . So one needed only to compute this ratio, and rank the values from lowest (ranking = 1) to highest (ranking = 4) ( 1 pt for each correct ranking or mass). Note that, although the ceramic ranks as the strongest of the four materials for its weight, its mode of failure at room temperature would be complete brittle fracture, without the yielding that the other materials would exhibit to indicate that they were being stressed beyond acceptable limits. For this reason, designers use generous safety factors of 100% or more when designing with brittle materials in loadbearing applications....
View
Full
Document
This document was uploaded on 11/08/2011.
 Spring '09

Click to edit the document details