Hw4s

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Unformatted text preview: HW# 4- solutions Q1) (a) E= 8 − 4 .6 3.4 = = 340MPa 0.025 − 0.015 0.01 (b) They are shown on the graph (c) Modulus of resilience = (11 − 0) × (0.035 − 0) × 1 = 0.1925MJ / m 3 2 (d) ν =− εy = εy εx Ly ⇒ ε y = −ν × ε x = −0.29 × 0.025 = −0.00725mm / mm ⇒ ∆L y = L y × ε y = 20 × −0.00725 = −0.145mm ∆L y Width of specimen at 8MPa = 20mm – 0.145mm=19.855 mm 1 τ(cw) σ(-) σ2 σ1 σ(+) τ(ccw) − 19.1 = 9.55 2 − 19.1 − 0 2 ) + 16.55 2 = 19.1 2 C= R= ( σ 1 = R − C = 19.1 − 9.55 = 9.55MPa σ 2 = − R − C =...
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This note was uploaded on 02/18/2009 for the course CE 231 taught by Professor Jasonweiss during the Spring '08 term at Purdue.

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