Exam2spring2004 - ENSC 3313 Spring 2004 TEST 2 KEY Name Answer all 10 questions show all work 2points each 1 The load-elongation plot for a tensile

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1 ENSC 3313, Spring 2004 TEST 2 - KEY Name_________________________ Answer all 10 questions, show all work, 2points each 1. The load-elongation plot for a tensile test specimen of gauge length 2 inches and diameter 0.5 inches is shown. Calculate the a) elastic modulus, b) yield strength, c) tensile strength, d) % elongation. The cross sectional area 2 2 2 . 0 4 ) 5 . 0 ( in A = π . Hence, stress=load/0.2 = 5 × load The gauge length = 2"; hence strain = elongation/2. a) E ~ (5 × 7000)/(0.01/2) ~ 7.0 × 10 6 psi b) YS ~ 5 × 9,000 ~ 45,000 psi c) TS ~ 5 × 14,000 ~ 70,000 psi d) % elongation ~ (0.092/2)100 ~ 4.6% 2. Shown is the plane (111) and the direction [ 01 1 ] in an aluminum FCC unit cell. Calculate the resolved shear stress in this slip system when 35MPa. is applied in the [001] direction. Recall: λ φ σ τ cos cos = r The resolved shear stress is given by the formula: cos cos = r Stress axis and plane normal: 3 1 cos = Stress axis and the slip direction: 2 1 cos = Resolved shear stress is: 408 . 0 6 = = r =14.28MPa x y z
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2 3. A 10 mm diameter 70 Cu- 30 Zn rod is to be drawn down to a 6mm diameter rod. The final product must have a tensile strength of above 450 MPa, a hardness of less than 75, and a % elongation (50 mm) of at least 15%. Specify a processing sequence.
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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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Exam2spring2004 - ENSC 3313 Spring 2004 TEST 2 KEY Name Answer all 10 questions show all work 2points each 1 The load-elongation plot for a tensile

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