Exam2-solutions

# Exam2-solutions - Midterm Exam II MASC 310 Oct 29, 2009 Dr....

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Midterm Exam II MASC 310 Oct 29, 2009 Dr. Carter CLOSED BOOK, CLOSED NOTES CALCULATORS ALLOWED __________Solution Set____________ Name

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Possibly helpful equations Mechanical Properties () ε σ + = + = 1 ln 1 T T Toughness = σ d ε Strengthening cos cos = λ φ τ R cos cos crss y = 2 / 1 + = d k y o yield He S y 1 ~ Failure 2 / 1 2 = a E s c π γ σ m =2 σ 0 a ρ t 1/2 max a Y K c c = Phase Diagrams Remember the Lever Rule! K t = σ max / σ o σ T = K T ( ) n o σ = F A o τ = F s A ε = δ L o σ = E ε ε ν = L ε x 100 L L L EL % o o f = o r t t E ) ( ) ( = Kt d d n o n =
1. Mechanical properties (6 points total) (A) Given the engineering stress-strain diagram below for a metallic material, estimate the (i) modulus of elasticity, (ii) tensile strength, (iii) percent elongation at failure and (iv) toughness (4 points) (i) The elastic modulus is just the slope of the initial linear portion of the curve or, from the inset: E = σ 2 −σ 1 ε 2 −ε 1 = (1300 0) MPa ( 6.25 x 10 3 0 ) = 210 × 10 3 MPa = 210 GPa (30.5 × 10 6 psi) The actual value is 207 GPa. Depending on “eyeball error” might get values between 190 and 230. Gpa. (ii) The tensile strength (the maximum on the curve) is approximately 1970 MPa (285,000 psi) (iii) The % elongation is 0.078 strain (~7.8%) (iv) The toughness is given by Toughness = σ d ε Estimating by eye the area under the full stress-strain curve gives ε σ d ~1700MPa * .078 ~ 1.3 × 10 8 J/m 3 Acceptable estimated values could range from 1 × 10 8 J/m 3 to 1.6 × 10 8 J/m 3 (B) On the basis of stress-strain behavior, polymers fall within three general classifications: brittle, plastic and highly elastic. Draw schematic stress strain curves for each, identifying failure and yield points for each (if applicable). (2 points) A=Brittle, B=plastic, C=highly elastic X= failure points Only B (plastic) exhibits significant yield, indicated by the arrow.

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2. Deformation and strengthening (12 points total) (A) Describe in your own words how the following strengthening techniques work to increase the yield strength of a metal. Be sure to explain how dislocations are involved in each technique and the dependence of each upon its relevant parameter. (i) Grain size reduction, (ii) Solid-solution strengthening, and (iii) Precipitation strengthening. (6 points) (i) Grain boundaries serve as barriers to dislocation motion. This is because the lattice orientation changes across the grain boundary, the resolved shear stress on the dislocation changes. Thus, refining the grain size of a polycrystalline material renders it harder and stronger. Yield stress will increase proportional to the inverse square root of the grain size d according to the Hall-Petch relationship, 2 / 1 + = d k y o yield σ . (ii) Solid-solution strengthening results from lattice strain interactions between impurity atoms and dislocations. The compressive and tensile stresses around the dislocation overlap with either compressive or tensile stresses induced by the impurity or alloy atoms, making it more difficult for the dislocation to move. As a result, yield stress will increase proportional to the inverse square
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## This note was uploaded on 10/11/2010 for the course ISE 460 taught by Professor Bottlik during the Spring '06 term at USC.

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Exam2-solutions - Midterm Exam II MASC 310 Oct 29, 2009 Dr....

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