Assignment6_solution

# Assignment6_solution - -2 and b = 0.286 nm Solution(a For...

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1 Assignment #6 Solutions ME264 Materials Science for Mechanical Engineering Spring 2009 Due Date: 08 April 2009 (4 pts) 1. (1pt) Explain what is meant by the ideal strength of a material. Solution: See section 9.2 from the textbook. 2. (2pts) (a) A polycrystalline aluminum allot contains a dispersion of hard particles of diameter 10 -8 m and average center-to-center spacing of 6 x 10 -8 m measured in the slip planes. Estimate their contribution to the tensile yield strength, σ y , of the alloy. (b) The alloy is used for the compressor blades of a small turbine. Adiabatic heating raises the blade temperature to 150 o C, and causes the particles to coarsen slowly. After 1000 hours they have grown to a diameter of 3 x 10 -8 m and are spaced 18 x 10 -8 m apart. Estimate the drop in yield strength. (The shear modulus of aluminum is 26 GN m

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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 m-Δ ≈ 2 3. (1pt) The yield stress of a sample of brass with a large grain size 20 MN m-2 . The yield stress of an otherwise identical sample with a grain size of 4 μ m was 120 MN m-2 . 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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Assignment6_solution - -2 and b = 0.286 nm Solution(a For...

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