TEST2_ENSC3313_SP2006_KEY

# TEST2_ENSC3313_SP200 - ENSC 3313 Spring 2006 TEST 2 KEY Name Answer all 10 questions show all work 2points each load(lb 1 Cartridge brass(C26000

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1 ENSC 3313, Spring 2006 TEST 2 - KEY Name_________________________ Answer all 10 questions, show all work, 2points each 1. Cartridge brass (C26000) has the following properties: E=16 × 10 6 psi, ν =0.35, TS=58 ksi, YS=46 ksi, and %elongation = 35%. Assume that you performed a tensile test on a C26000 sample with a 0.25 in diameter and 2 in gage length. Label the load elongation curve (not stress-strain) with the relevant values. Cross sectional area is: 05 . 0 4 ) 25 . 0 ( A 2 = π = Yield load = 46,000*0.05 = 2300 lb Max load = 58,000*0.05 = 2900 lb Yield strain = YS/E = 46/16 × 10 -3 = 2.9 × 10 -3 Yield elongation = 2*2.9 × 10 -3 = 5.8 × 10 -3 in Failure elongation = 0.35*2 = 0.7 in 2. Calculate the resolved shear stress ( τ r ) relevant for dislocation motion in the primary slip system of a copper crystal (FCC) if a tensile stress of 30 ksi is applied in the (001) direction. Recall the cosine of the angle between two vectors is: 2 1 2 1 v v v v cos = θ The primary slip systems in FCC are the {111} planes and <110> directions. As sketched, the relevant plane and direction are (111) and [ 01 1] 3 1 cos ; 2 1 cos = φ = λ Hence, ksi 2 . 12 6 r = σ = τ elongation (in) 0.7 0.0058 2900 2300 load (lb) x y z (111) [101] [001]

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## This note was uploaded on 01/10/2012 for the course ENSC 3313 taught by Professor Staff during the Fall '11 term at Oklahoma State.

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TEST2_ENSC3313_SP200 - ENSC 3313 Spring 2006 TEST 2 KEY Name Answer all 10 questions show all work 2points each load(lb 1 Cartridge brass(C26000

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