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Unformatted text preview: 2 , and b = 0.286 nm.) Solution: (a) For the alloy 8 10 9 2 8 2 2 5 10 , 2.86 10 , 26 10 So 1.5 10 . But 3 for polycrystals. Hence 450 . y y y y L m b m G N m Gb N m L MN m τ σ= × = × = × = = × = ≈ (b) New 8 15 10 . L m= × So 7 2 4.96 10 . y Gb N m L= = × 2 3 149 y y y MN m= ≈ Drop in yield strength 2 300 y MN mΔ ≈ 2 3. (1pt) The yield stress of a sample of brass with a large grain size 20 MN m2 . The yield stress of an otherwise identical sample with a grain size of 4 μ m was 120 MN m2 . Why did the yield stress increase in this way? What is the value of β in equation (10.5) for the brass? Solution: ( ) 6 1/2 1/2 2 1/2 2 3/2 1/2 4 10 500 100 20 500 100 0.2 500 d m d m MN m m MN m MN m m β= × == ⋅ = =...
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 Spring '09
 Asst.Prof.A.FethiOkyar
 Mechanical Engineering, Tensile strength, yield strength, tensile yield strength

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